Hc Verma I Solutions for Class 11 Science Physics Chapter 5 Newton S Laws Of Motion are provided here with simple step-by-step explanations. These solutions for Newton S Laws Of Motion are extremely popular among Class 11 Science students for Physics Newton S Laws Of Motion Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Hc Verma I Book of Class 11 Science Physics Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Hc Verma I Solutions. All Hc Verma I Solutions for class Class 11 Science Physics are prepared by experts and are 100% accurate.

#### Page No 76:

#### Answer:

No, the accelerating elevator will affect the weight of both sides of the beam balance. So, the net effect of the accelerating elevator cancels out, and we get the actual mass.

#### Page No 76:

#### Answer:

During free fall:

Acceleration of the boy = Acceleration of mass *M* =* g *

Acceleration of mass *M* w.r.t. boy, *a* = 0

So, the force exerted by the box on the boy's head = *M × a* = 0

Yes, the force greatly increases during the period he balances himself after striking the ground because of the weight of the box.

#### Page No 76:

#### Answer:

(a) In the car, the path of the coin will be vertically downward because the only force acting on the coin is gravity in the downward direction.

(b) In a free falling elevator, the coin as well as the person will be in a condition of weightlessness. So, the coin will remain stationary w.r.t. the person.

#### Page No 76:

#### Answer:

If no force acts on the particle it cannot change its direction. So, it is not possible for a particle to describe a curved path if no force acts on it.

Yes, the answer depends on the frame of reference chosen to view the particle if the frame of reference describes a curved path.

#### Page No 76:

#### Answer:

We are pushed forward because of the inertia of motion, as our body opposes the sudden change.

#### Page No 76:

#### Answer:

We can't find a body whose acceleration is zero with respect to all other bodies in the universe because every body in the universe is moving with respect to other bodies.As we live on earth which itself is accelerates due to its revolution around the sun and spinning about its own axis, so whatever observations and measurements ,we make , are w.r.t to earth which itself is not an inertial frame.Similarly all other planets are also in motion around the sun so tdeally no inertial frame is possible.

#### Page No 76:

#### Answer:

Yes, if the force on the object is zero, its acceleration w.r.t. all the other objects will we zero. So, the frame will necessarily be an inertial frame.

#### Page No 76:

#### Answer:

(a)

The reading of the balance = Tension in the string

And tension in the string = 20*g*

So, the reading of the balance = 20*g** =* 200 N

(b) If the balance is heavy, the reading will not change because the weight of spring balance does not affect the tension in the string.

(c) If the blocks have unequal masses, the spring balance will accelerate towards the heavy block with an acceleration *a. *Then the reading will be equal to the tension in the string.

Suppose *m*_{1}_{ > }*m*_{2}.

Then tension in the string,

$T=\frac{2{m}_{1}{m}_{2}g}{{m}_{1}+{m}_{2}}$

#### Page No 77:

#### Answer:

No. The acceleration of the particle can also be zero if the vector sum of all the forces is zero, i.e. no net force acts on the particle.

#### Page No 77:

#### Answer:

The force of friction acting between my feet and ground is responsible for my deceleration.

#### Page No 77:

#### Answer:

In both the cases, change in momentum is same but the time interval during which momentum changes to zero is less in the first case. So, by $F=\frac{dP}{dt}$ , force in the first case will be more. That's why we are injured when we jump barefoot on a hard surface.

#### Page No 77:

#### Answer:

The forces on the rope must be equal and opposite, according to Newton's third law. But not all the forces acting on each team are equal. The friction between one team and the ground does not depend on the other team and can be larger on one side than on the other. In addition, the grips on the rope need not be equal and opposite. Thus, the net force acting on each team from all sources need not be equal.

#### Page No 77:

#### Answer:

Air applies a velocity-dependent force on the parachute in upward direction when the parachute opens. This force opposes the gravitational force acting on the spy. Hence, the net force in the downward direction decreases and the spy decelerates.

#### Page No 77:

#### Answer:

Yes, this is an example of Newton's third law of motion, which sates that every action has an equal and opposite reaction.

#### Page No 77:

#### Answer:

Yes, the forces due to the spring on the two blocks are equal and opposite.

But it's not an example of Newton's third law because there are three objects (2 blocks + 1 spring). Spring force on one block and force by the same block on the spring is an action-reaction pair.

#### Page No 77:

#### Answer:

No, w.r.t. the ground frame, the person's head is not really pushed backward.

As the train moves, the lower portion of the passenger's body starts moving with the train, but the upper portion tries to be in rest according to Newton's first law and hence, the passenger seems to be pushed backward.

#### Page No 77:

#### Answer:

No, a person sitting inside the compartment can't tell just by looking at the plumb line whether the train is accelerating on a horizontal straight track or moving on an incline.

When the train is accelerating along the horizontal, the tension in the string is $m\sqrt{{g}^{2}+{a}^{2}}$; when it is moving on the inclined plane, the tension is* mg*. So, by measuring the tension in the string we can differentiate between the two cases.

#### Page No 77:

#### Answer:

(c) *w*_{1}* + *w_{2}

From the free-body diagram,

(*w*_{1} + *w*_{2}) – *N *= 0

*N = **w*_{1}* + **w*_{2}

The ceiling pulls the chain by a force (*w*_{1} + *w*_{2}).

#### Page No 77:

#### Answer:

(b) the ground on the horse

The horse pushes the ground in the backward direction and, in turn, the ground pushes the horse in the forward direction, according to Newton's third law of motion.

#### Page No 77:

#### Answer:

(d) the road

The car pushes the ground in the backward direction and according to the third law of motion, reaction force of the ground in the forward direction acts on the car.

#### Page No 77:

#### Answer:

(a) Both the scales will read 10 kg.

From the free-body diagram,

*K*_{1}*x*_{1} = *mg* = 10$\times $9.8 = 98 N

*K*_{2}*x*_{2} = *K*_{1}*x*_{1}

So, *K*_{1}*x*_{1} = *K*_{2}*x*_{2} = 98 N

Therefore, both the spring balances will read the same mass, i.e. 10 kg.

#### Page No 77:

#### Answer:

(c) *mg* cosθ

From the free-body diagram,

*N = mg *cosθ

Normal force exerted by the plane on the block is *mg* cosθ.

#### Page No 77:

#### Answer:

(b) *mg*/cosθ

Free-body Diagram of the Small Block of Mass '*m*'

The block is at equilibrium w.r.t. to wedge. Therefore,

*mg* sin*θ* = *ma *cosθ

*⇒ a = **g*tan*θ*

Normal reaction on the block is

*N = mg *cos*θ** *+* **ma *sin*θ*

Putting the value of *a,* we get:

*N = mg *cos*θ** + mg *tan*θ*sin*θ*

*$N=mg\mathrm{cos}\theta +mg\frac{\mathrm{sin}\theta}{\mathrm{cos}\theta}\mathrm{sin}\theta N=\frac{mg}{\mathrm{cos}\theta}$*

#### Page No 77:

#### Answer:

(d) remain standing

If the earth suddenly stops attracting objects placed near its surface, the net force on the person will become zero and according to the first law of motion, the person will remain standing.

#### Page No 77:

#### Answer:

(a) zero

Using, *F*_{net}_{ }=* ma*,

*a* = 0 *⇒ **F*_{net}_{ = 0}

As the whole system is at rest, the resultant force on the charged particle at A is zero.

#### Page No 78:

#### Answer:

(b) *F*_{1} may be equal to *F*_{2}.

Any force applied in the direction opposite the motion of the particle decelerates it to rest.

#### Page No 78:

#### Answer:

(b) *A* will go higher than *B*.

Let the air exert a constant resistance force = F (in downward direction).

Acceleration of particle A in downward direction due to air resistance, *a*_{A} = *F*/*m*_{A}.

Acceleration of particle B in downward direction due to air resistance,* **a*_{B} = *F*/*m*_{B}.

*m*_{A} > *m*_{B}

*a*_{A} < *a*_{B}

$S=ut+\frac{1}{2}a{t}^{2}$

$\mathrm{So},{H}_{A}=ut-\frac{1}{2}({a}_{A}+g){t}^{2}$

${H}_{B}=ut-\frac{1}{2}({a}_{B}+g){t}^{2}$

${H}_{A}>{H}_{B}$

Therefore, *A* will go higher than *B. *

#### Page No 78:

#### Answer:

(c) remain at the top of the wedge

Downward gravitational force will be balanced by the upward pseudo force (because of the motion of the wedge in downward direction). The block will remain at its position, as both the box and the inclined plane are falling with the same acceleration (*g*).

#### Page No 78:

#### Answer:

(b) *t*_{1} > *t*_{2}

Let acceleration due to air resistance force be *a. *

Let *H* be maximum height attained by the particle.

Direction of air resistance force is in the direction of motion.

In the upward direction of motion, ${a}_{\mathrm{eff}}=\left|g-a\right|$.

${t}_{1}=\sqrt{\frac{2H}{\left|g-a\right|}}...\left(1\right)$

In the downward direction of motion, ${a}_{\mathrm{eff}}=g+a$.

${t}_{2}=\sqrt{\frac{2H}{g+a}}...\left(2\right)$

So, *t*_{1} > *t*_{2}.

#### Page No 78:

#### Answer:

(a) *t*_{1} = *t*_{2}

After the coin is dropped, the only force acting on it is gravity, which is same for both the cases.

So *t*_{1} = *t*_{2}.

#### Page No 78:

#### Answer:

(c) *x*

The moving train does not put any extra force on the alpha particle and the recoiling nucleus. So, the distance between the alpha particle and the recoiling nucleus at a time *t* after the decay, as measured by the passenger, will be same as before, i.e.* x*.

#### Page No 78:

#### Answer:

(b) cannot be an inertial frame because the earth is revolving around the sun

(d) cannot be an inertial frame because the earth is rotating about its axis

A reference frame attached to the earth cannot be an inertial frame because the earth is revolving around the sun and also rotating about its axis.

#### Page No 78:

#### Answer:

(c) the frame may be inertial but the resultant force on the particle is zero

(d) the frame may be non-inertial but there is a non-zero resultant force

According to Newton's second law which says that net force acting on the particle is equal to rate of change of momentum ( or mathematically F = ma), so if a particle is at rest then *F*_{net} = ma = m$\frac{dv}{dt}=m\frac{d\left(0\right)}{dt}=m\times 0=0$.

Now, if the frame is inertial, then the resultant force on the particle is zero.

If the frame is non-inertial,

vector sum of all the forces plus a pseudo force is zero.

i.e. *F*_{net} ≠ 0.

#### Page No 78:

#### Answer:

(a) Both the frames are inertial.

(b) Both the frames are non-inertial.

*S*_{1} is moving with constant velocity w.r.t frame *S*_{2}. So, if *S*_{1}_{ }is inertial, then *S*_{2}_{ }will be inertial and if *S*_{1}_{ }is non-inertial, then *S*_{2}_{ }will be non-inertial.

#### Page No 78:

#### Answer:

(a) *AB*

(c) *CD*

Slope of the *x-t* graph gives velocity. In the regions *AB* and *CD,* slope or velocity is constant, i.e. acceleration is zero. Hence, from the second law, force is zero in these regions.

#### Page No 78:

#### Answer:

(a) *F*_{1}_{ }= *F*_{2} = *F*, for *t* < 0

At *t *< 0, the block is in equilibrium in the horizontal direction.

So, *F*_{1}_{ }= *F*_{2} = *F*

At *t* > 0, *F*_{2}_{ }= 0 and *F*_{1}_{ }= *F*.

#### Page No 79:

#### Answer:

(b) going up and speeding up

(c) going down and slowing down

It means normal force exerted by the floor of the elevator on the person is greater that the weight of the person.

i.e. *N* > *mg*

(i) Going up and speeding up:

*a*_{eff} =* g + a*

*N = **ma*_{eff} = *mg + ma *(*N > mg*)

(ii) Going down and speeding up:

*a*_{eff}_{ }= *g – a*

*N = mg – ma *(*N < mg*)

(iii) Going down and slowing down:

*a*_{eff} = *g – *(*–a*)* = g + a*

*N = mg + ma *(*N > mg*)

(iv) Going up and slowing down:

*a*_{eff} = *g – a*

* N = mg – ma *(*N < mg*)

#### Page No 79:

#### Answer:

(c) going up with uniform speed

(d) going down with uniform speed

Tension in the cable = Weight of the elevator

Or, total upward force = total downward force

That is, there's no acceleration or uniform velocity.

So, the elevator is going up/down with uniform speed.

#### Page No 79:

#### Answer:

(d) *F*_{1} = 0, *F*_{2} = 0

${a}_{{s}_{2}{s}_{1}}=a...\left(1\right)$

Acceleration of the particle w.r.t. to *S*_{1}* = **F*_{1}*/m*

Acceleration of the particle w.r.t. to* S*_{2}* = **F*_{2}*/m*

If we assume *F*_{1} = 0 and *F*_{2} = 0,

we can conclude that ${a}_{{s}_{2}{s}_{1}}=0...\left(2\right)$

From equations (1) and (2), we can say that our assumption is wrong.

And *F*_{1} = 0, *F*_{2} = 0 is not possible.

#### Page No 79:

#### Answer:

(d) He might have used a non-inertial frame

If no force is acting on a particle and yet, its acceleration is non-zero, it means the observer is in a non-inertial frame.

#### Page No 79:

#### Answer:

Given:

Mass of the block,* m* = 2 kg,

Distance covered, *S* = 10 m and initial velocity, *u* = 0

Let* a* be the acceleration of the block.

Using, $\mathrm{S}=ut+\frac{1}{2}a{t}^{2}$, we get:

$10=\frac{1}{2}a\left({2}^{2}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 10=2a\phantom{\rule{0ex}{0ex}}\Rightarrow a=5\mathrm{m}/{\mathrm{s}}^{2}$

∴ Force,* F* = *ma *= 2 × 5 = 10 N

#### Page No 79:

#### Answer:

Given:

Initial speed of the car, *u* = 40 km/hr $=\frac{4000}{3600}=11.11\mathrm{m}/\mathrm{s}$

Final speed of the car, *v* = 0

Mass of the car,* m* = 2000 kg

Distance to be travelled by the car before coming to rest, *s** *= 4m

Acceleration, $a=\frac{{v}^{2}-{u}^{2}}{2\mathrm{s}}$

$\Rightarrow a=\frac{{0}^{2}-{\left(11.11\right)}^{2}}{2\times 4}=\frac{-123.43}{8}=-15.42\mathrm{m}/{\mathrm{s}}^{2}$

∴ Average force to be applied to stop the car, *F* = *ma*

*⇒ F* = 2000 × 15.42 ≈ 3.1 × 10^{4} N

#### Page No 79:

#### Answer:

Initial velocity of the electrons is negligible, i.e. *u* = 0.

Final velocity of the electrons*, v* = 5 × 10^{6} m/s

Distance travelled by the electrons,

*s* = 1 cm = 1 × 10^{−2} m

∴ Acceleration, $a=\frac{{v}^{2}-{u}^{2}}{2\mathrm{S}}$

$\Rightarrow a=\frac{{\left(5\times {10}^{5}\right)}^{2}-0}{2\times 1\times {10}^{-3}}=\frac{25\times {10}^{12}}{2\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}\Rightarrow a=12.5\times {10}^{14}\mathrm{m}/{\mathrm{s}}^{2}$

So, force on the electrons, *F* = *ma*

*⇒* *F* = 9.1 × 10^{−31} × 12.5 × 10^{−14}

= 1.1 × 10^{−15} N

#### Page No 79:

#### Answer:

The free-body diagrams for both the blocks are shown below:

From the free-body diagram of the 0.3 kg block,

*T* = 0.3*g*

⇒ *T*= 0.3 × 10 = 3 N

Now, from the free-body diagram of the 0.2 kg block,

*T*_{1}_{ }= 0.2*g* + *T*

⇒ *T*_{1}= 0.2 × 10 + 3 = 5 N

∴ The tensions in the two strings are 5 N and 3 N, respectively.

#### Page No 79:

#### Answer:

Let* a *be the common acceleration of the blocks.

For block 1,

$F-T=ma$ ...(1)

For block 2,

*T* = *ma* ...(2)

Subtracting equation (2) from (1), we get:

$F-2T=0$

$\Rightarrow T=\frac{F}{2}$

#### Page No 79:

#### Answer:

Given:

Mass of the particle,* m* = 50 g = 5 × 10^{−2} kg

Slope of the* v-t* graph gives acceleration.

At *t* = 2 s,

Slope = $\frac{15}{3}=5\mathrm{m}/{\mathrm{s}}^{2}$

So, acceleration, *a* = 5 m/s^{2}

*F* = *ma* = 5 × 10^{−2} × 5

⇒ *F* = 0.25 N along the motion.

At *t *= 4 s,

Slope = 0

So, acceleration,* a* = 0

⇒ *F* = 0

At *t* = 6 sec,

Slope = $\frac{-15}{3}=-5\mathrm{m}/{\mathrm{s}}^{2}$

So, acceleration, *a* = − 5 m/s^{2}

*F* = *ma* = − 5 × 10^{−2} × 5

⇒ *F* = − 0.25 N along the motion

or, *F *= 0.25 N opposite the motion.

#### Page No 79:

#### Answer:

Let *F' *= force exerted by the experimenter on block A and *F* be the force exerted by block A on block B.

Let *a* be the acceleration produced in the system.

For block A,

$F\text{'}-F={m}_{A}a$ ...(1)

For block B,

*F* = *m*_{B}*a** ...*(2)

Dividing equation (1) by (2), we get:

$\frac{F\text{'}}{F}-1=\frac{{m}_{A}}{{m}_{B}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow F\text{'}=F\left(1+\frac{{m}_{\mathrm{A}}}{{m}_{\mathrm{B}}}\right)$

∴ Force exerted by the experimenter on block A is $F\left(1+\frac{{m}_{\mathrm{A}}}{{m}_{\mathrm{B}}}\right)$.

#### Page No 79:

#### Answer:

Given:

Radius of a raindrop, *r* = 1 mm = 10^{−3} m

Mass of a raindrop, *m* = 4 *mg* = 4 × 10^{−6} kg

Distance coved by the drop on the head, *s* = 10^{−3} m

Initial speed of the drop, *v* = 0

Final speed of the drop, *u* = 30 m/s

Using $a=\frac{{v}^{2}-{u}^{2}}{2s}$, we get:

$a=\frac{-{\left(30\right)}^{2}}{2\times {10}^{-3}}=-4.5\times {10}^{5}m/{s}^{2}$

Force, *F = ma
⇒ F = *4 × 10

^{−6}× 4.5 × 10

^{5}

= 1.8 N

#### Page No 79:

#### Answer:

Displacement of the particle from the mean position*, x* = 20 cm = 0.2 m

*k* = 15 N/m

Mass of the particle, *m* = 0.3 kg

Acceleration, $a=\frac{\left|\mathit{F}\right|}{m}$

$\Rightarrow a=\frac{kx}{m}=\frac{15\left(0.2\right)}{0.3}=\frac{3}{0.3}=10\mathrm{m}/{\mathrm{s}}^{2}$

So, the initial acceleration when the particle is released from a point *x* = 20 cm is 10 m/s^{2}.

#### Page No 79:

#### Answer:

Let the block *m* be displaced towards left by displacement *x*.

∴ *F*_{1} = $-$*k*_{1}*x* (compressed)

*F*_{2} = $-$*k*_{2}*x* (expanded)

$ma={F}_{1}+{F}_{2}$

$\Rightarrow a=\frac{-x\left({k}_{1}+{k}_{2}\right)}{m}$

i.e. $\left({k}_{1}+{k}_{2}\right)\frac{x}{m}$ opposite the displacement or towards the mean position.

#### Page No 79:

#### Answer:

Mass of block A, *m* = 5 kg

*F *= *ma* = 10 N

$\Rightarrow a=\frac{10}{5}=2\mathrm{m}/{\mathrm{s}}^{2}$

As there is no friction between A and B, when block A moves, block B remains at rest in its position.

Initial velocity of A, *u* = 0

Distance covered by A to separate out,

*s* = 0.2 m

Using $s=ut+\frac{1}{2}a{t}^{2}$, we get:

$0.2=0+\frac{1}{2}\times 2{t}^{2}$

⇒ *t*^{2} = 0.2

⇒* t* = 0.44 s ≈ 0.45 s

#### Page No 79:

#### Answer:

(a) At any depth, let the ropes makes an angle *θ* with the vertical.

From the free-body diagram,

*F*cos*θ** + **F*cos*θ** *− *mg *= 0

2*F*cos*θ** = mg*

$\Rightarrow F=\frac{mg}{2cos\theta}$

As the man moves up, *θ* increases, i.e. cos*θ* decreases. Thus,* F* increases.

(b) When the man is at depth *h*,

$\mathrm{cos}\theta =\frac{h}{\sqrt{{\left(d/2\right)}^{2}+{h}^{2}}}$

$F=\frac{mg}{2\mathrm{cos}\theta}\phantom{\rule{0ex}{0ex}}=\frac{mg}{2\left[h/\sqrt{{\left({\displaystyle \frac{d}{2}}\right)}^{2}+{h}^{2}}\right]}\phantom{\rule{0ex}{0ex}}=\frac{mg}{4h}\sqrt{{d}^{2}+4{h}^{2}}$

#### Page No 80:

#### Answer:

When the elevator is descending, a pseudo-force acts on it in the upward direction, as shown in the figure.

From the free-body diagram of block A,

$mg-N=ma$

$N=m\left(g-a\right)\phantom{\rule{0ex}{0ex}}\Rightarrow N=0.5\left(10-2\right)=4\mathrm{N}\phantom{\rule{0ex}{0ex}}$

So, the force exerted by the block A on the block B is 4 N.

#### Page No 80:

#### Answer:

(a) When the elevator goes up with acceleration 1.2 m/s^{2}:

$T=mg+ma$

*⇒ T* = 0.05 (9.8 + 1.2) = 0.55 N

(b) Goes up with deceleration 1.2 m/s^{2} :

$T=mg+m\left(-a\right)=m\left(g-a\right)$

⇒ *T* = 0.05 (9.8 − 1.2) = 0.43 N

(c) Goes up with uniform velocity:

$T=mg$

⇒ *T* = 0.05 × 9.8 = 0.49 N

(d) Goes down with acceleration 1.2 m/s^{2} :

$T+ma=mg\phantom{\rule{0ex}{0ex}}\Rightarrow T=m\left(g-a\right)$

⇒ *T* = 0.05 (9.8 − 1.2) = 0.43 N

(e) Goes down with deceleration 1.2 m/s^{2} :

$T+m\left(-a\right)=mg\phantom{\rule{0ex}{0ex}}\Rightarrow T=m\left(g+a\right)$

⇒ *T *= 0.05 (9.8 + 1.2) = 0.55 N

(f) Goes down with uniform velocity:

$T=mg$

⇒* T *= 0.05 × 9.8 = 0.49 N

#### Page No 80:

#### Answer:

Maximum weight will be recorded when the elevator accelerates upwards.

Let *N* be the normal reaction on the person by the weighing machine.

So, from the free-body diagram of the person,

$N=mg+ma$ ...(1)

This is maximum weight, *N* = 72 × 9.9 N

When decelerating upwards, minimum weight will be recorded.

$N\text{'}=mg+m\left(-a\right)$ ...(2)

This is minimum weight, *N*' = 60 × 9.9 N

From equations (1) and (2), we have:

2 *mg *= 1306.8

$\Rightarrow m=\frac{1306.8}{2\times 9.9}=66\mathrm{kg}$

So, the true mass of the man is 66 kg.

And true weight = 66 $\times $ 9.9 = 653.4 N

(b) Using equation (1) to find the acceleration, we get:

*mg* + *ma* = 72 × 9.9

$\Rightarrow a=\frac{72\times 9.9-66\times 9.9}{66}=\frac{9.9\times 6}{66}=\frac{9.9}{11}\phantom{\rule{0ex}{0ex}}\Rightarrow a=0.9\mathrm{m}/{\mathrm{s}}^{2}$

#### Page No 80:

#### Answer:

Let the left and right blocks be A and B, respectively.

And let the acceleration of the 3 kg mass relative to the elevator be '*a*' in the downward direction.

From the free-body diagram,

${m}_{A}a=T-{m}_{A}g-\frac{{m}_{A}g}{10}...\left(1\right)$

${m}_{B}a={m}_{B}g+\frac{{m}_{B}g}{10}-T...\left(2\right)$

Adding both the equations, we get:

$a\left({m}_{A}+{m}_{B}\right)=\left({m}_{B}-{m}_{A}\right)g+\left({m}_{B}-{m}_{A}\right)\frac{g}{10}$

Putting value of the masses,we get:

$9a=\frac{33g}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{a}{g}=\frac{11}{30}...\left(3\right)$

Now, using equation (1), we get:

$T={m}_{A}\left(a+g+\frac{g}{10}\right)$

The reading of the spring balance = $\frac{2T}{g}=\frac{2}{g}{m}_{A}\left(a+g+\frac{g}{10}\right)$

$\Rightarrow 2\times 1.5\left(\frac{a}{g}+1+\frac{1}{10}\right)=3\left(\frac{11}{30}+1+\frac{1}{10}\right)\phantom{\rule{0ex}{0ex}}=4.4\mathrm{kg}$

#### Page No 80:

#### Answer:

Given,

mass of the first block, *m* = 2 kg

*k* = 100 N/m

Let elongation in the spring be *x. *

From the free-body diagram,

*kx* = *mg*

$x=\frac{mg}{k}=\frac{2\times 9.8}{100}\phantom{\rule{0ex}{0ex}}=\frac{19.6}{100}=0.196\approx 0.2\mathrm{m}$

Suppose, further elongation, when the 1 kg block is added, is *$\u2206x$*.

Then,$k\left(x+\u2206x\right)=m\text{'}g$

⇒ *k$\u2206x$*= 3*g* − 2*g* = *g*

$\Rightarrow \u2206x=\frac{g}{k}=\frac{9.8}{100}=0.098\approx 0.1\mathrm{m}$

#### Page No 80:

#### Answer:

When the ceiling of the elevator is going up with an acceleration 'a', then a pseudo-force acts on the block in the downward direction.

*
a* = 2 m/s

^{2}

From the free-body diagram of the block,

*kx = m*

*g*+

*ma*

⇒

*kx*= 2

*g*+ 2

*a*

= 2 × 9.8 + 2 × 2

= 19.6 + 4

$\Rightarrow x=\frac{23.6}{100}=0.236\approx 0.24\mathrm{m}$

When 1 kg body is added,

total mass = (2 + 1) kg = 3 kg

Let elongation be

*x*'.

∴

*kx'*= 3

*g*+ 3

*a*= 3 × 9.8 + 6

$\Rightarrow x\text{'}=\frac{35.4}{100}\phantom{\rule{0ex}{0ex}}=0.354\approx 0.36\mathrm{m}$

So, further elongation =

*x' − x*

= 0.36 − 0.24 = 0.12 m.

#### Page No 80:

#### Answer:

Let *M* be mass of the balloon.

Let the air resistance force on balloon be *F *.

Given that* F ∝ v*.

⇒ *F _{ }= kv*,

where

*k*= proportionality constant.

When the balloon is moving downward with constant velocity,

*B + kv = Mg*...(i)

$\Rightarrow M=\frac{B+kv}{g}$

Let the mass of the balloon be

*M' so that it can*rise with a constant velocity

*v*in the upward direction.

*B = Mg + kv*

$\Rightarrow M\text{'}=\frac{B+kv}{g}$

∴ Amount of mass that should be removed =

*M − M'.*

$\u2206M=\frac{B+kv}{g}-\frac{B-kv}{g}\phantom{\rule{0ex}{0ex}}=\frac{B+kv-B+kv}{g}\phantom{\rule{0ex}{0ex}}=\frac{2kv}{g}=\frac{2\left(Mg-B\right)}{g}\phantom{\rule{0ex}{0ex}}=2\left\{M-\frac{B}{g}\right\}$

$\u2206M=\frac{B+kv}{g}-\frac{B-kv}{g}\phantom{\rule{0ex}{0ex}}=\frac{B+kv-B+kv}{g}\phantom{\rule{0ex}{0ex}}=\frac{2kv}{g}=\frac{2\left(Mg-B\right)}{g}\phantom{\rule{0ex}{0ex}}=2\left\{M-\frac{B}{g}\right\}$

#### Page No 80:

#### Answer:

Let *U* be the upward force of water acting on the plastic box.

Let *m* be the initial mass of the plastic box.

When the empty plastic box is accelerating upward,

$U-mg=\frac{mg}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow U=\frac{7mg}{6}$

$\Rightarrow m=\frac{6U}{7g}....\left(i\right)$

Let M be the final mass of the box after putting some sand in it.

$Mg-U=\frac{Mg}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow Mg-\frac{Mg}{6}=U\phantom{\rule{0ex}{0ex}}\Rightarrow M=\frac{6U}{5g}....\left(ii\right)$

Mass added$=\frac{6U}{5g}-\frac{6U}{7g}$

$=\frac{6U\left(7-5\right)}{35g}\phantom{\rule{0ex}{0ex}}=\frac{6U\xb72}{35g}$

From equation (i), $m=\frac{6U}{7g}$

∴ Mass added$=\frac{2}{5}m$ .

#### Page No 80:

#### Answer:

For the particle to move without being deflected and with constant velocity, the net force on the particle should be zero.

$\overrightarrow{F}+m\overrightarrow{g}=0$

$\Rightarrow \left(\overrightarrow{v}\times \overrightarrow{A}\right)+\overrightarrow{mg}=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\overrightarrow{v}\times \overrightarrow{A}\right)=-\overrightarrow{mg}$

$\left|vA\mathrm{sin}\theta \right|=\left|mg\right|$

$\therefore v=\frac{mg}{A\mathrm{sin}\theta}$

*v* will be minimum when sin*θ* = 1.

⇒ *θ* = 90°

$\therefore {v}_{min}=\frac{mg}{A}$

#### Page No 80:

#### Answer:

The masses of the blocks are *m*_{1} = 0.3 kg and *m*_{2} = 0.6 kg

The free-body diagrams of both the masses are shown below:

For mass *m*_{1},

* T* − *m*_{1}*g* = *m*_{1}*a* ...(i)

For mass *m*_{2},

*m*_{2}*g* − *T*= *m*_{2}*a* ...(ii)

Adding equations (i) and (ii), we get:

*g*(*m*_{2} − *m*_{1}) = *a*(*m*_{1} + *m*_{2})

$\Rightarrow a=g\left(\frac{{m}_{2}-{m}_{1}}{{m}_{1}+{m}_{2}}\right)\phantom{\rule{0ex}{0ex}}=9.8\times \frac{0.6-0.3}{0.6+0.3}\phantom{\rule{0ex}{0ex}}=3.266\mathrm{m}/{\mathrm{s}}^{2}$

(a) *t* = 2 s, *a* = 3.266 ms^{−2}, *u* = 0

So, the distance travelled by the body,

$\mathrm{S}=ut+\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}=0+\frac{1}{2}\left(3.266\right){2}^{2}=6.5\mathrm{m}$

(b) From equation (i),

*T* = *m*_{1} (*g *+ *a*)

= 0.3 (3.8 + 3.26) = 3.9 N

(c)The force exerted by the clamp on the pulley,

*F = *2*T** *= 2 × 3.9 = 7.8 N

#### Page No 80:

#### Answer:

*a* = 3.26 m/s^{2}, *T* = 3.9 N

After 2 s, velocity of mass *m*_{1}_{,}

*v* = *u* + *at* = 0 + 3.26 × 2

= 6.52 m/s upward

At this time, *m*_{2} is moving 6.52 m/s downward.

At time 2 s, *m*_{2} stops for a moment. But *m*_{1} is moving upward with velocity 6.52 m/s. It will continue to move till final velocity (at highest point) becomes zero.

Here, *v* = 0, *u* = 6.52 m/s

*a* = −*g* = − 9.8 m/s^{2}

*v* = *u* + *at* = 6.52 + (−9.8)*t*

$\Rightarrow t=\frac{6.52}{9.8}\approx \frac{2}{3}\mathrm{sec}$

After this time, the mass* **m*_{1} also starts moving downward.

So, the string becomes tight again after $\frac{2}{3}\mathrm{s}$.

#### Page No 80:

#### Answer:

Mass per unit length$=\frac{3}{30}\mathrm{kg}/\mathrm{cm}$ = 0.10 kg/cm

Mass of the 10 cm part, *m*_{1} = 1 kg

Mass of the 20 cm part, *m*_{2} = 2 kg

Let *F* = contact force between them.

From the free-body diagram,

${m}_{1}a=F-20...\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}{m}_{2}a=32-F...\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Adding}\mathrm{both}\mathrm{the}\mathrm{equation}\text{s},\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}a=\frac{12}{{m}_{1}+{m}_{2}}=\frac{12}{3}=4\mathrm{m}/{\mathrm{s}}^{2}$

So, contact force,

*F* = 20 + 1*a*

*F* = 20 + 4 = 24 N

#### Page No 81:

#### Answer:

Mass of each block is 1 kg, $\mathrm{sin}{\mathrm{\theta}}_{1}=\frac{4}{5}$, $\mathrm{sin}{\mathrm{\theta}}_{2}=\frac{3}{5}$.

The free-body diagrams for both the boxes are shown below:

*m**g*sin*θ*_{1} − *T = ma* ...(i)

* T − **mg*sin*θ*_{2} = *ma *...(ii)

Adding equations (i) and (ii),we get:

*mg*(sin*θ*_{1} *− *sin*θ*_{2}) = 2*ma** *

⇒ 2*a* = *g* (sin*θ*_{1} − sin*θ*_{2})

$\Rightarrow a=\frac{g}{5}\times \frac{1}{2}\phantom{\rule{0ex}{0ex}}=\frac{g}{10}$

#### Page No 81:

#### Answer:

The free-body diagrams for both the blocks are shown below:

From the free-body diagram of block of mass *m*_{1},

*m*_{1}*a* =* T − F* ...(i)

From the free-body diagram of block of mass *m*_{2},

*m*_{2}*a* = *m*_{2}*g* − T ...(ii)

Adding both the equations, we get:

$a\left({m}_{1}+{m}_{2}\right)={m}_{2}g-\frac{{m}_{2}g}{2}\left[\mathrm{because}F=\frac{{m}_{2}g}{2}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow a=\frac{{m}_{2}g}{2\left({m}_{1}+{m}_{2}\right)}$

So, the acceleration of mass *m*_{1},

$a=\frac{{m}_{2}g}{2\left({m}_{1}+{m}_{2}\right)},\mathrm{towards}\text{the}\mathrm{right}.$

#### Page No 81:

#### Answer:

Let the acceleration of the blocks be *a*.

The free-body diagrams for both the blocks are shown below:

From the free-body diagram,

*m*_{1}*a* = *m*_{1}*g* +* F − T ...*(i)

Again, from the free-body diagram,

*m*_{2}*a* = *T − m*_{2}*g* *− F ...*(ii)

Adding equations (i) and (ii), we have:

$a=g\frac{{m}_{1}-{m}_{2}}{{m}_{1}+{m}_{2}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow a=\frac{3g}{7}=\frac{29.4}{7}\phantom{\rule{0ex}{0ex}}=4.2\mathrm{m}/{\mathrm{s}}^{2}$

Hence, acceleration of the block is 4.2 m/s^{2}.

After the string breaks, *m*_{1} moves downward with force *F* acting downward. Then,

*m*_{1}*a* = *F* + *m*_{1}*g*

5*a* = 1 + 5*g*

$\Rightarrow a=\frac{5g+1}{5}\phantom{\rule{0ex}{0ex}}=g+0.2\mathrm{m}/{\mathrm{s}}^{2}$

#### Page No 81:

#### Answer:

The free-body diagram for mass *m*_{1} is shown below:

(Figure 1)

The free-body diagram for mass *m*_{2} is shown below:

(Figure 2)

The free-body diagram for mass *m*_{3} is shown below:

(Figure 3)

Suppose the block *m*_{1} moves upward with acceleration *a*_{1} and the blocks *m*_{2} and *m*_{3} have relative acceleration *a*_{2} due to the difference of weight between them.

So, the actual acceleration of the blocks *m*_{1}, *m*_{2} and *m*_{3} will be *a*_{1}, (*a*_{1} − *a*_{2}) and (*a*_{1} + *a*_{2}), as shown.

From figure 2, *T* − 1*g* − 1*a*_{1} = 0 ...(i)

From figure 3, $\frac{T}{2}-2g-2\left({a}_{1}-{a}_{2}\right)=0...\left(\mathrm{ii}\right)$

From figure 4, $\frac{T}{2}-3g-3\left({a}_{1}+{a}_{2}\right)=0...\left(\mathrm{iii}\right)$

From equations (i) and (ii), eliminating T, we get:

1*g* + 1*a*_{2} = 4*g* + 4 (*a*_{1} + *a*_{2})

5*a*_{2} − 4*a*_{1} = 3*g* ...(iv)

From equations (ii) and (iii), we get:

2*g* + 2(*a*_{1} − *a*_{2}) = 3*g* − 3 (*a*_{1} − *a*_{2})

5*a*_{1} + *a*_{2} = *g* ...(v)

Solving equations (iv) and (v), we get:

${a}_{1}=\frac{2g}{29}\phantom{\rule{0ex}{0ex}}\mathrm{and}{a}_{2}=g-5{a}_{1}\phantom{\rule{0ex}{0ex}}\Rightarrow {a}_{2}=g-\frac{10g}{29}=\frac{19g}{29}\phantom{\rule{0ex}{0ex}}\mathrm{So},{a}_{1}-{a}_{2}=\frac{2g}{29}-\frac{19g}{29}=-\frac{17g}{29}\phantom{\rule{0ex}{0ex}}\mathrm{and}{a}_{1}+{a}_{2}=\frac{2g}{29}+\frac{19g}{29}=\frac{21g}{29}$

So, accelerations of *m*_{1}, *m*_{2} and *m*_{3} are $\frac{19g}{29}\left(\mathrm{up}\right),\frac{17g}{29}\left(\mathrm{down}\right)\mathrm{and}\frac{21g}{29}\left(\mathrm{down}\right)$ , respectively.

Now, *u* = 0, *s* = 20 cm = 0.2 m

${a}_{2}=\frac{19g}{29}\phantom{\rule{0ex}{0ex}}\therefore s=ut+\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 0.2=\frac{1}{2}\times \frac{19}{29}g{t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow t=0.25\mathrm{s}$

#### Page No 81:

#### Answer:

For m_{1} to be at rest, *a*_{1}_{ }= 0.

*T *−* **m*_{1}*g* = 0

*T = **m*_{1}*g** *...(i)

For mass *m*_{2},

*T*/2 − 2*g** *= 2*a** *

*T =** *4*a* + 4*g** ** * ...(ii)

For mass *m*_{3},

3*g* –* T*/2= 2*a** *

*T =** *6*g* − 6*a** ** * ...(ii)

From equations (ii) and (iii), we get:

3*T** *– 12*g** *= 12*g* – 2*T** *

*T* = 24g/5= 4.08g

Putting the value of *T* in equation (i), we get:

*m*_{1}_{ }= 4.8kg

#### Page No 81:

#### Answer:

The free-body diagrams for both the bodies are shown below:

*T* + *ma* =*mg *...(i)

and *T* = *ma* ...(ii)

From equations (i) and (ii), we get:

*ma* + *ma* = m*g*

⇒ 2*ma* = *g*

$\Rightarrow a=\frac{g}{2}=\frac{10}{5}=5\mathrm{m}/{\mathrm{s}}^{2}$

From equation (ii),

*T* = *ma* = 5 N

#### Page No 81:

#### Answer:

Let the acceleration of mass *M* be* a*.

So, the acceleration of mass 2*M** *will be* $\frac{a}{2}$.*

(a) 2*M*(*a*/2) − 2*T* = 0

⇒* Ma = *2*T*

*T + Ma − Mg *= 0

$\Rightarrow \frac{Ma}{2}+Ma=Mg\phantom{\rule{0ex}{0ex}}\Rightarrow 3Ma=2Mg\phantom{\rule{0ex}{0ex}}\Rightarrow a=\frac{2g}{3}$

(b) Tension, $T=\frac{Ma}{2}=\frac{M}{2}\times \frac{2g}{3}=\frac{Mg}{3}$

(c) Let *T*' = resultant of tensions

$\therefore T\text{'}=\sqrt{{T}^{2}+{T}^{2}}=\sqrt{2}T\phantom{\rule{0ex}{0ex}}\therefore T\text{'}=\sqrt{2}T=\frac{\sqrt{2}Mg}{3}\phantom{\rule{0ex}{0ex}}\mathrm{Again},\mathrm{tan}\theta =\frac{T}{T}=1\phantom{\rule{0ex}{0ex}}\Rightarrow \theta =45\xb0$

So, the force exerted by the clamp on the pulley is $\frac{\sqrt{2}\mathrm{M}g}{3}$ at an angle of 45° with the horizontal.

#### Page No 81:

#### Answer:

The free-body diagram of the system is shown below:

Let acceleration of the block of mass 2*M* be *a*.

So, acceleration of the block of mass *M* will be 2*a**.*

*M*(2*a*) + *Mg*sin*θ** *−* T* = 0

⇒ *T* = 2*Ma* + *Mg*sin*θ* ...(i)

2*T* + 2*Ma* − 2*Mg* = 0

From equation (i),

2(2*Ma* + *Mg*sin*θ*) + 2*Ma* − 2*M*g = 0

4*Ma* + 2*Mg*sin*θ* + 2*Ma* −* Mg* = 0

6*Ma* + 2*Mg*sin30° + 2*Mg* = 0

6*Ma** = Mg*

$\Rightarrow a=\frac{g}{6}$

Hence, the acceleration of mass $M=2a=2\times \frac{g}{6}=\frac{g}{3}\left(\mathrm{up}\mathrm{the}\mathrm{plane}\right).$

#### Page No 82:

#### Answer:

The free-body diagram of the system is shown below:

Block ‘*m*’ will have the same acceleration as that of *M',* as it does not slip over *M*'.

From the free body diagrams,

*T + Ma – Mg = *0* ...*(i)

*T – M'a – **R*sin*θ** = *0 ...(ii)

*R*sin*θ** – ma* = 0

*R*cos*θ** – mg *= 0

Eliminating

*T, R*and

*a*from the above equations, we get:

$M=\frac{M\mathit{\text{'}}\mathit{+}m}{\mathrm{cot}\mathrm{\theta}-1}$

#### Page No 82:

#### Answer:

(a) 5*a* + *T* − 5*g* = 0

From free-body diagram (1),

*T* = 5*g* − 5*a* .....(i)

Again, $\left(\frac{1}{2}\right)T-4g-8a=0$

⇒ *T* − 8*g* − 16*a* = 0

(1)

From free-body diagram (2),

*T* = 8*g* + 16*a* ......(ii)

From equations (i) and (ii), we get:

5*g* − 5*a* = 8*g* + 16*a*

$\Rightarrow 21a=-3g-a=-\frac{9}{7}$

So, the acceleration of the 5 kg mass is $\frac{9}{7}\mathrm{m}/{\mathrm{s}}^{2}\left(\mathrm{upward}\right)$ and that of the 4 kg mass is $\mathit{2}a\mathit{=}\frac{\mathit{2}\mathit{g}}{\mathit{7}}\mathit{}\left(\mathrm{downward}\right)$.

(b) From free body diagram-3,

$4a-\frac{T}{2}=0$

(2)

⇒ 8*a* − *T* = 0

⇒ *T* = 8*a*

Again, *T *+ 5*a* − 5*g* = 0

From free body diagram-4,

8*a* + 5*a* − 5*g* = s0

⇒ 13*a* − 5*g** *= 0

$\Rightarrow a=\frac{5g}{13}\left(\mathrm{downward}\right)$

Acceleration of mass 2 kg is $2a=\frac{10}{13}\left(g\right)$ and 5 kg is $\frac{5g}{13}$.

(c) *T* + 1*a* − 1*g* = 0

From free body diagram-5

*T* = 1*g* − 1*a* .....(i)

Again, from free body diagram-6,

$\frac{T}{2}-2g-4a=0$

⇒ *T *− 4*g* − 8*a* = 0 .....(ii)

From equation (i)

1*g* − 1*a* − 4*g* − 8*a* = 0

$\Rightarrow a=\frac{g}{3}\left(\mathrm{downward}\right)$

Acceleration of mass 1 kg is $\frac{g}{3}\left(\mathrm{upward}\right)$,

Acceleration of mass 2 kg is $\frac{2g}{3}\left(\mathrm{downward}\right)$.

(3)

#### Page No 82:

#### Answer:

Given,

*m*_{1} = 100 g = 0.1 kg

*m*_{2} = 500 g = 0.5 kg

*m*_{3} = 50 g = 0.05 kg

The free-body diagram for the system is shown below:

From the free-body diagram of the 500 g block,

*T* + 0.5*a* − 0.5*g* = 0 .....(i)

From the free-body diagram of the 50 g block,

*T*_{1} + 0.05*g* − 0.05*a* = *a* ....(ii)

From the free-body diagram of the 100 g block,

T_{1}_{ }+ 0.1*a* − *T* + 0.5*g* = 0 ....(iii)

From equation (ii),

*T*_{1} = 0.05*g* + 0.05*a* .....(iv)

From equation (i),

*T*_{1} = 0.5*g* − 0.5*a* .....(v)

Equation (iii) becomes

*T*_{1} + 0.1*a* − *T* + 0.05*g* = 0

From equations (iv) and (v), we get:

0.05*g* + 0.05*a* + 0.1*a* − 0.5*g* + 0.5*a* + 0.05*g* = 0

0.65*a* = 0.4 *g*

$\Rightarrow a=\frac{0.4}{0.65}g\phantom{\rule{0ex}{0ex}}=\frac{40}{65}g=\frac{8}{13}g\left(\mathrm{downward}\right)$

So, the acceleration of the 500 gm block is $\frac{8g}{13}$downward.

#### Page No 82:

#### Answer:

Mass of the monkey, *m* = 15 kg,

Acceleration of the monkey in the upward direction, *a* = 1 m/s^{2}

The free-body diagram of the monkey is shown below:

From the free-body diagram,

*T* − [15*g* + 15(*a*)] = 0

*T* − [15*g* + 15(1)] = 0

⇒ *T* = 5 (10 + 1)

⇒ *T* = 15 × 11 = 165 N

The monkey should apply a force of 165 N to the rope.

Initial velocity, *u* = 0

*s* = 5 m

Using, $s=ut+\frac{1}{2}a{t}^{2}$, we get:

$5=0+\left(\frac{1}{2}\right)\times 1\times {t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {t}^{2}=5\times 2\phantom{\rule{0ex}{0ex}}\Rightarrow t=\sqrt{10}\mathrm{s}$

Hence, the time required to reach the ceiling is $\sqrt{10}\mathrm{s}$.

#### Page No 82:

#### Answer:

Suppose the monkey accelerates upward with acceleration *a* and the block accelerates downward with acceleration *a*'.

Let force exerted by the monkey be *F*.

From the free-body diagram of the monkey, we get:

*F*− *mg* − *ma* = 0 ...(i)

*⇒ **F* = *mg* + *ma*

Again, from the free-body diagram of the block,

*F* + *ma*' − *mg* = 0

*mg* + *ma* + *ma'* − *mg* = 0 [From (i)]

*⇒ ma* = −*ma*'

*⇒ a* = −*a*'

If acceleration −*a*' is in downward direction then the acceleration *a*' will be in upward direction.

This implies that the block and the monkey move in the same direction with equal acceleration.

If initially they were at rest (no force is exerted by the monkey), then their separation will not change as time passes because both are moving same direction with equal acceleration.

#### Page No 82:

#### Answer:

Let the acceleration of monkey A upwards be *a*, so that a maximum tension of 30 N is produced in its tail.

(i)

(ii)

*T* − 5*g* − 30 − 5*a* = 0 ...(i)

30 − 2*g* − 2*a* = 0 ...(ii)

From equations (i) and (ii), we have:

* T* = 105 N (max.)

and *a* = 5 m/s^{2}

So, A can apply a maximum force of 105 N on the rope to carry monkey B with it.

For minimum force, there is no acceleration of A and B.

*T*_{1} = weight of monkey B

⇒ *T*_{1} = 20 N

Rewriting equation (i) for monkey A, we get:

*T* − 5*g* − 20 = 0

⇒ *T* = 70 N

∴ To carry monkey B with it, monkey A should apply a force of magnitude between 70 N and 105 N.

#### Page No 82:

#### Answer:

(i) Given, mass of the man = 60 kg

Let *W*' = apparent weight of the man in this case

From the free-body diagram of the man,

*W*' + *T* − 60*g* = 0

⇒ *T* = 60*g* − *W*' ...(i)

From the free-body diagram of the box,

*T* − *W*' = 30*g* = 0 ...(ii)

From equation (i), we get:

60*g* − *W*' − *W*' − 30*g* = 0

⇒* W*' = 15*g*

Hence, the weight recorded on the machine is 15 kg.

(ii) To find his actual weight, suppose the force applied by the men on the rope is *T *because of which the box accelerates upward with an acceleration *a*'.Here we need to find ${T}^{\text{'}}$.

Correct weight = *W* = 60*g*

From the free-body diagram of the man,

T' + *W* − 60*g* − 60*a* = 0

⇒ T' − 60*a* = 0

⇒ T' = 60*a** *...(i)

From the free-body diagram of the box,

T' − *W* − 30*g* − 30*a* = 0

⇒ T' − 60*g* − 30*g* − 30*a* = 0

⇒ T' = 30*a* − 900 ...(ii)

From equations (i) and (ii), we get:

T' = 2T' − 1800

T' = 1800 N

So, the man should exert a force of 1800 N on the rope to record his correct weight on the machine.

#### Page No 83:

#### Answer:

The force on the block which makes the body move down the plane is the component of its weight parallel to the inclined surface.

*F* = *mg *sin*θ*

Acceleration, *g* = sin θ

Initial velocity of block, *u* = 0

Distance to be covered

*s* = *l*

*a* = *g* sin θ

Using, $s=ut+\frac{1}{2}a{t}^{2}$, we get:

$l=0+\frac{1}{2}\left(g\mathrm{sin}\theta \right){t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {t}^{2}=\frac{2l}{g\mathrm{sin}\theta}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Time}\mathrm{taken},t=\sqrt{\frac{2\mathrm{l}}{\mathrm{gsin}\theta}}$

#### Page No 83:

#### Answer:

Let the pendulum (formed by the ball and the string) make angle *θ * with the vertical.

From the free-body diagram,

*T*cos*θ** − mg = 0*

*T*cos*θ** = mg*

$\Rightarrow T=\frac{mg}{\mathrm{cos}\mathrm{\theta}}...\left(\mathrm{i}\right)$

And, *ma* − *T* sin θ = 0

*⇒ ma* = T sin θ

$\Rightarrow T=\frac{ma}{\mathrm{sin}\mathrm{\theta}}....\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\mathrm{\theta}=\frac{a}{g}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{\theta}={\mathrm{tan}}^{-1}\frac{a}{g}$

So, the angle formed by the ball with the vertical is ${\mathrm{tan}}^{-1}\left(\frac{a}{g}\right)$ .

(ii) Let the angle of the incline be *θ*.

From the diagram,

⇒* ma* cos θ = *mg* sin θ

$\frac{\mathrm{sin}\mathrm{\theta}}{\mathrm{cos}\mathrm{\theta}}=\frac{a}{g}\phantom{\rule{0ex}{0ex}}\mathrm{tan}\mathrm{\theta}=\frac{a}{g}\mathrm{\theta}={\mathrm{tan}}^{-1}\left(\frac{a}{g}\right)$

So, the angle of incline is ${\mathrm{tan}}^{-1}\left(\frac{a}{g}\right)$.

#### Page No 83:

#### Answer:

The free-body diagram of the system is shown below:

The two bodies are separated because the elevator is moving downward with an acceleration of 12 m/s^{2} (>g) and the body moves with acceleration, *g* = 10 m/s^{2} [Freely falling body]

Now, for the block:

*g* = 10 m/s^{2}*, u* = 0*, t* = 0.2 s

So, the distance travelled by the block is given by

$s=ut+\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}=0+\frac{1}{2}10\times {\left(0.2\right)}^{2}=5\times 0.04\phantom{\rule{0ex}{0ex}}=0.2\mathrm{m}=20\mathrm{cm}$

The displacement of the body is 20 cm during the first 0.2 s.

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