3 .   A   r e c tan g u l a r   s o l i d   b l o c k   h a s   s i d e s   4 × 10 × 20   c m   a n d   a   d e n s i t y   o f   7 . 5   g   c m - 1 . i f   i t   r e s t s   o n   a   h o r i z o n t a l   f l a t   s u r f a c e   c a l c u l a t e   t h e   m i n i m u m   a n d   m a x i m u m   p r e s s u r e   i t   c a n   e x e r t i n   S I   u n i t s   .   A l s o   c a l c u l a t e   t h e   t h r u s t   i n   b o t h   c a s e s .         A n s :   3000   P a   a n d   15000   P a   ,   60   N

Dear Student
Volume Of the block = 4*10*20 = 800 cm3
Mass of the block m = volume * density = 800 * 7.5 = 6000 gm = 6 kg
Weight of the block W = mg = 6 * 10 = 60N
Thrust by the surface should be equal to the weight because it is at equilibrium so net force should be zero.
Therefore thrust in both the cases = 60N

Pressure exerted by the block  P = Thrust / face area
Maximum pressure Pmax = W/Amin = 60/(4*10*10-4) = 15000 Pa
Minimum Pressure PMin =  W/Amax = 60/(20*10*10-4) = 3000 Pa

I hope the solution is helpful.
Keep asking question, keep learning.
Regards



 

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