Question

A 50 kg block rests on four cylindrical rubber blocks. Each block has a height of 4 cm and a cross-sectional area of $16{cm}^{2}$. The shear modulus of rubber is $2 \times {10}^{6} {N}/{{m}^{2}}$. A sideways force of 500N is applied to the motor. The distance that the motor moves sideways is
A.0.156 cm
B.1.56 cm
C.0.312 cm
D.0.204 cm

Answer 1

Hint: To solve this problem, first find the force applied on each block by taking the ratio of total force applied to the motor and number of cylindrical blocks. Then, use the formula for shear modulus which is also known as modulus of rigidity. Substitute the given values and the force applied on each block in the formula for shear modulus. Thus, calculate the value of x which gives the distance that motor moves sideways.
Formula used:
$F= \dfrac {{F}_{0}}{n}$
$ \eta = \dfrac {Fh}{Ax}$

Complete step by step answer:

Given: Mass of block (M)= 50 Kg
            Number of cylindrical blocks (n)= 4
            Height of cylindrical block (h)= 4 cm= 0.04 m
            Cross-sectional area of cylindrical block (A)= $16{cm}^{2}=16 \times {10}^{-4}{m}^{2}$
            Shear modulus of rubber $\eta$= $2 \times {10}^{6}N/{{m}^{2}} $
            Force (${F}_{0}$)= 500 N
Force applied on each block is given by,
$F= \dfrac {{F}_{0}}{n}$
Substituting the values we get,
$F = \dfrac {500}{4}$
$\Rightarrow F = 125 N$
Shear modulus is given by, $ \eta = \dfrac {Fh}{Ax}$
Where, x is the change in distance
Substituting the values in above equation we get,
$2 \times {10}^{6} = \dfrac {125 \times 0.04}{16\times {10}^{-4} \times x}$
Rearranging the above equation we get,
$x= \dfrac {125 \times 0.04}{16 \times {10}^{-4} \times 2 \times {10}^{6}}$
$\Rightarrow x= 15.625 \times {10}^{-4} m$
$\Rightarrow x= 0.156 cm$
Hence, the distance that the motor moves sideways is 0.156 cm.

So, the correct answer is option A i.e. 0.156 cm.

Note:
Students must remember to convert the units. So, convert the units either to CGS or MKS. Not converting the units will give you an incorrect answer. Students must also remember that shear modulus exists in solid only. If students do not remember the formula for shear modulus, they can derive it by taking the ratio of shear stress and shear strain. Shear modulus is associated with the change in the shape of the body.