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Graduate Aptitude Test in Engineering

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1

MCQ (More than One Correct Answer)

Two solid spheres A and B of equal volumes but of different densities d_{A} and d_{B} are connected by a string. They are fully immersed in a fluid of density d_{F}. They get arranged into an equilibrium state as shown in the figure with a tension in the string. The arrangement is possible only if

A

d_{A} < d_{F}

B

d_{B} > d_{F}

C

d_{A} > d_{F}

D

d_{A} + d_{B} = 2d_{F}

Let V be the volume of each sphere and let T be the tension in the string.

Buoyant force on sphere A is U_{A} = d_{F}Vg

Buoyant force on sphere B is U_{B} = d_{F}Vg

Weight of A is W_{A} = d_{A}Vg

Weight of B is W_{B} = d_{B}Vg

The free body diagrams of A and B are as follows. (see Fig. 11.34)

For equilibrium,

U_{A} = T + W_{A} and U_{B} + T = W_{B}

i.e. d_{F}Vg = T + d_{A}Vg ..... (i)

and d_{F}Vg + T = d_{B}Vg .... (ii)

From Eq. (i) $${d_F} = {T \over {Vg}} + {d_A}$$. Hence, d_{F} > d_{A}. So choice (a) is correct.

From Eq. (ii) $${d_B} = {T \over {Vg}} + {d_F}$$. Hence d_{B} > d_{F}. So choice (b) is also correct.

Eliminating T from Eqs. (i) and (ii) we get $$2{d_F} = {d_A} + {d_B}$$, which is choice (d).

So, the correct choices are (a), (b) and (d).

2

MCQ (More than One Correct Answer)

A composite block is made of slabs A, B, C, D and E of different thermal conductivities (given in terms of a constant K) and sizes (given in terms of length, L) as shown in the figure. All slabs are of same width. Heat Q flows only from left to right through the blocks. Then, in steady-state

A

heat flow through A and E slabs are same.

B

heat flow through slab E is maximum.

C

temperature difference across slab E is smallest.

D

heat flow through C = heat flow through B + heat flow through D.

Let T_{1}, T_{2}, T_{3} and T_{4} be the temperatures at slab interfaces as shown in the figure.

The rate of heat flow across a slab, with thermal conductivity $$\kappa $$, width w, thickness l, length x, and temperature difference $$\Delta$$T, is given by

$$dQ/dt = \kappa lw(\Delta T/x)$$.

Thus, the heat flow rates through the given slabs are

$${{d{Q_A}} \over {dt}} = {{(2K)(4Lw)({T_1} - {T_2})} \over L} = 8Kw({T_1} - {T_2})$$ ...... (1)

$${{d{Q_B}} \over {dt}} = {{(3K)(Lw)({T_2} - {T_3})} \over {4L}} = {3 \over 4}Kw({T_2} - {T_3})$$ ...... (2)

$${{d{Q_C}} \over {dt}} = {{(4K)(2Lw)({T_2} - {T_3})} \over {4L}} = 2Kw({T_2} - {T_3})$$ ...... (3)

$${{d{Q_D}} \over {dt}} = {{(5K)(Lw)({T_2} - {T_3})} \over {4L}} = {5 \over 4}Kw({T_2} - {T_3})$$ ..... (4)

$${{d{Q_E}} \over {dt}} = {{(6K)(4Lw)({T_3} - {T_4})} \over L} = 24Kw({T_3} - {T_4})$$ ..... (5)

In the steady state, heat flow into the system is equal to the heat flow out of the system i.e., $$d{Q_A}/dt = d{Q_E}/dt$$.

Use equations (1) and (5) to get

$${T_1} - {T_2} = 3({T_3} - {T_4})$$ ..... (6)

Similarly, the steady state condition for the interface at temperature T_{2} is

$$d{Q_A}/dt = d{Q_B}/dt + d{Q_C}/dt + d{Q_D}/dt$$,

which gives (by using equations (1)-(4))

$$2({T_1} - {T_2}) = {T_2} - {T_3}$$ ..... (7)

Use equations (6) and (7) to get

$${T_3} - {T_4} = (1/3)({T_1} - {T_2}) = (1/6)({T_2}/{T_3})$$.

Also, equations (2)-(4) give $$d{Q_C}/dt = d{Q_B}/dt + d{Q_D}/dt$$.

On those following papers in MCQ (Multiple Correct Answer)

Number in Brackets after Paper Indicates No. of Questions

JEE Advanced 2022 Paper 2 Online (1)

JEE Advanced 2022 Paper 1 Online (1)

JEE Advanced 2021 Paper 1 Online (1)

JEE Advanced 2020 Paper 1 Offline (1)

JEE Advanced 2019 Paper 1 Offline (1)

JEE Advanced 2018 Paper 2 Offline (1)

JEE Advanced 2018 Paper 1 Offline (1)

JEE Advanced 2015 Paper 2 Offline (3)

JEE Advanced 2014 Paper 1 Offline (1)

JEE Advanced 2013 Paper 1 Offline (1)

IIT-JEE 2012 Paper 1 Offline (1)

IIT-JEE 2011 Paper 2 Offline (1)

IIT-JEE 2011 Paper 1 Offline (1)

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