Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

Given in the figure are two blocks $$A$$ and $$B$$ of weight 20 N and 100 N, respectively. These are being pressed against a wall by a force $$F$$ as shown. If the coefficient of friction between the blocks is 0.1 and between block $$B$$ and the wall is 0.15, the frictional force applied by the wall on block $$B$$ is :

A

$$120$$ $$N$$

B

$$150$$ $$N$$

C

$$100$$ $$N$$

D

$$80$$ $$N$$

Assuming both the blocks are stationary

From the F.B.D of both the blocks we can find,

$$N=F$$

f

f

Considering those two blocks as one system and due to equilibrium $$f=120N$$

2

MCQ (Single Correct Answer)

A block of mass $$m$$ is placed on a surface with a vertical cross section given by $$y = {{{x^3}} \over 6}.$$ If the coefficient of friction is $$0.5,$$ the maximum height above the ground at which the block can be placed without slipping is:

A

$${1 \over 6}m$$

B

$${2 \over 3}m$$

C

$${1 \over 3}m$$

D

$${1 \over 2}m$$

At limiting equilibrium, $$\mu = \tan \theta $$

Equation of the surface,

$$y = {{{x^3}} \over 6}$$

Slope, $$\tan \theta = \mu = {{dy} \over {dx}} = {{{x^2}} \over 2}$$

Given that, Coefficient of friction $$\mu = 0.5$$

$$\therefore$$ $$\,\,\,\,0.5 = {{{x^2}} \over 2}$$

$$ \Rightarrow \,\,\,x = \pm \,1$$

Now, $$y = {{{x^3}} \over 6} = {1 \over 6}m$$

Equation of the surface,

$$y = {{{x^3}} \over 6}$$

Slope, $$\tan \theta = \mu = {{dy} \over {dx}} = {{{x^2}} \over 2}$$

Given that, Coefficient of friction $$\mu = 0.5$$

$$\therefore$$ $$\,\,\,\,0.5 = {{{x^2}} \over 2}$$

$$ \Rightarrow \,\,\,x = \pm \,1$$

Now, $$y = {{{x^3}} \over 6} = {1 \over 6}m$$

3

MCQ (Single Correct Answer)

Two fixed frictionless inclined planes making an angle $${30^ \circ }$$ and $${60^ \circ }$$ with the vertical are shown in the figure. Two blocks $$A$$ and $$B$$ are placed on the two planes. What is the relative vertical acceleration of $$A$$ with respect to $$B$$ ?

A

$$4.9m{s^{ - 2}}$$ in horizontal direction

B

$$9.8m{s^{ - 2}}$$ in vertical direction

C

Zero

D

$$4.9m{s^{ - 2}}$$ in vertical direction

Along inclined plane the equation of motion of the body

$$mg\,\sin \,\theta = ma$$ $$\,\,\,\,\,\,\,\,$$ $$\therefore$$ $$a = g\,\sin \,\theta $$

where $$a$$ is along the inclined plane.

$$\therefore$$ vertical component of acceleration is $$\left( {g\sin \theta } \right)\sin \theta $$ = $$g\,{\sin ^2}\theta $$ (Along vertical)

For block A,

$${a_{A\left( {along\,vertical} \right)}} = g{\sin ^2}60^\circ $$

For block B,

$${a_{B\left( {along\,vertical} \right)}} = g{\sin ^2}30^\circ $$

$$\therefore$$ relative vertical acceleration of $$A$$ with respect to $$B$$ is

$$g{\sin ^2}60^\circ $$ - $$g{\sin ^2}30^\circ $$

=$$g\left( {{{\sin }^2}60 - {{\sin }^2}\left. {30} \right]} \right.$$

$$= 4.9$$ $$\,\,m/{s^2}$$ in vertical direction

$$mg\,\sin \,\theta = ma$$ $$\,\,\,\,\,\,\,\,$$ $$\therefore$$ $$a = g\,\sin \,\theta $$

where $$a$$ is along the inclined plane.

$$\therefore$$ vertical component of acceleration is $$\left( {g\sin \theta } \right)\sin \theta $$ = $$g\,{\sin ^2}\theta $$ (Along vertical)

For block A,

$${a_{A\left( {along\,vertical} \right)}} = g{\sin ^2}60^\circ $$

For block B,

$${a_{B\left( {along\,vertical} \right)}} = g{\sin ^2}30^\circ $$

$$\therefore$$ relative vertical acceleration of $$A$$ with respect to $$B$$ is

$$g{\sin ^2}60^\circ $$ - $$g{\sin ^2}30^\circ $$

=$$g\left( {{{\sin }^2}60 - {{\sin }^2}\left. {30} \right]} \right.$$

$$= 4.9$$ $$\,\,m/{s^2}$$ in vertical direction

4

MCQ (Single Correct Answer)

A block of mass $$m$$ is connected to another block of $$mass$$ $$M$$ by a spring (massless) of spring constant $$k.$$ The block are kept on a smooth horizontal plane. Initially the blocks are at rest and the spring is unstretched. Then a constant force $$F$$ starts acting on the block of mass $$M$$ to pull it. Find the force of the block of mass $$m.$$

A

$${{MF} \over {\left( {m + M} \right)}}$$

B

$${{mF} \over M}$$

C

$${{\left( {M + m} \right)F} \over m}$$

D

$${{mF} \over {\left( {m + M} \right)}}$$

we get $$T = ma$$

we get $$F-T=Ma$$

where $$T$$ is force due to spring

$$ \Rightarrow F - ma = Ma$$

$$ \Rightarrow$$ $$F=Ma+ma$$

$$\therefore$$ $$a = {F \over {M + m}}$$

Now, force acting on the block of mass $$m$$ is

$$ma = m\left( {{F \over {M + m}}} \right) = {{mF} \over {m + M}}.$$

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