Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

A bob of mass 'm' suspended by a thread of length l undergoes simple harmonic oscillations with time period T. If the bob is immersed in a liquid that has density $${1 \over 4}$$ times that of the bob and the length of the thread is increased by 1/3^{rd} of the original length, then the time period of the simple harmonic oscillations will be :-

A

T

B

$${3 \over 2}$$T

C

$${3 \over 4}$$T

D

$${4 \over 3}$$T

$$T = 2\pi \sqrt {l/g} $$

When bob is immersed in liquid

mg_{eff} = mg $$-$$ Buoyant force

mg_{eff} = mg $$-$$ v$$\sigma$$g ($$\sigma$$ = density of liquid)

$$ = mg - v{\rho \over 4}g$$

$$ = mg - {{mg} \over 4} = {{3mg} \over 4}$$

$$\therefore$$ $${g_{eff}} = {{3g} \over 4}$$

$${T_1} = 2\pi \sqrt {{{{l_1}} \over {{g_{eff}}}}} $$

$${l_1} = l + {l \over 3} = {{4l} \over 3},\,{l_{eff}} = {{3g} \over 4}$$

By solving

$${T_1} = {4 \over 3}2\pi \sqrt {l/g} $$

$${T_1} = {{4T} \over 3}$$

When bob is immersed in liquid

mg

mg

$$ = mg - v{\rho \over 4}g$$

$$ = mg - {{mg} \over 4} = {{3mg} \over 4}$$

$$\therefore$$ $${g_{eff}} = {{3g} \over 4}$$

$${T_1} = 2\pi \sqrt {{{{l_1}} \over {{g_{eff}}}}} $$

$${l_1} = l + {l \over 3} = {{4l} \over 3},\,{l_{eff}} = {{3g} \over 4}$$

By solving

$${T_1} = {4 \over 3}2\pi \sqrt {l/g} $$

$${T_1} = {{4T} \over 3}$$

2

MCQ (Single Correct Answer)

Four identical hollow cylindrical columns of mild steel support a big structure of mass 50 $$\times$$ 10^{3} kg. The inner and outer radii of each column are 50 cm and 100 cm respectively. Assuming uniform local distribution, calculate the compression strain of each column. [Use Y = 2.0 $$\times$$ 10^{11} Pa, g = 9.8 m/s^{2}]

A

3.60 $$\times$$ 10^{$$-$$8}

B

2.60 $$\times$$ 10^{$$-$$7}

C

1.87 $$\times$$ 10^{$$-$$3}

D

7.07 $$\times$$ 10^{$$-$$4}

Force on each column = $${{mg} \over 4}$$

Strain = $${{mg} \over {4AY}}$$

$$ = {{50 \times {{10}^3} \times 9.8} \over {4 \times \pi (1 - 0.25) \times 2 \times {{10}^{11}}}}$$

= 2.6 $$\times$$ 10^{$$-$$7}

Strain = $${{mg} \over {4AY}}$$

$$ = {{50 \times {{10}^3} \times 9.8} \over {4 \times \pi (1 - 0.25) \times 2 \times {{10}^{11}}}}$$

= 2.6 $$\times$$ 10

3

MCQ (Single Correct Answer)

A uniform heavy rod of weight 10 kg ms^{$$-$$2}, cross-sectional area 100 cm^{2} and length 20 cm is hanging from a fixed support. Young modulus of the material of the rod is 2 $$\times$$ 10^{11} Nm^{$$-$$2}. Neglecting the lateral contraction, find the elongation of rod due to its own weight.

A

2 $$\times$$ 10^{$$-$$9} m

B

5 $$\times$$ 10^{$$-$$8} m

C

4 $$\times$$ 10^{$$-$$8} m

D

5 $$\times$$ 10^{$$-$$10} m

We know,

$$\Delta l = {{WL} \over {2AY}}$$

$$\Delta l = {{10 \times 1} \over {2 \times 5}} \times 100 \times {10^{ - 4}} \times 2 \times {10^{11}}$$

$$\Delta l = {1 \over 2} \times {10^{ - 9}} = 5 \times {10^{ - 10}}$$ m

Option (d)

4

MCQ (Single Correct Answer)

In Millikan's oil drop experiment, what is viscous force acting on an uncharged drop of radius 2.0 $$\times$$ 10^{$$-$$5} m and density 1.2 $$\times$$ 10^{3} kgm^{$$-$$3} ? Take viscosity of liquid = 1.8 $$\times$$ 10^{$$-$$5} Nsm^{$$-$$2}. (Neglect buoyancy due to air).

A

3.8 $$\times$$ 10^{$$-$$11} N

B

3.9 $$\times$$ 10^{$$-$$10} N

C

1.8 $$\times$$ 10^{$$-$$10} N

D

5.8 $$\times$$ 10^{$$-$$10} N

Viscous force = Weight

$$ = \rho \times \left( {{4 \over 3}\pi {r^3}} \right)g$$

= 3.9 $$\times$$ 10^{$$-$$10}

$$ = \rho \times \left( {{4 \over 3}\pi {r^3}} \right)g$$

= 3.9 $$\times$$ 10

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