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### IIT-JEE 2011 Paper 2 Offline

MCQ (More than One Correct Answer)

Two solid spheres A and B of equal volumes but of different densities dA and dB are connected by a string. They are fully immersed in a fluid of density dF. 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
dA < dF
B
dB > dF
C
dA > dF
D
dA + dB = 2dF

## Explanation

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

Buoyant force on sphere A is UA = dFVg

Buoyant force on sphere B is UB = dFVg

Weight of A is WA = dAVg

Weight of B is WB = dBVg

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

For equilibrium,

UA = T + WA and UB + T = WB

i.e. dFVg = T + dAVg ..... (i)

and dFVg + T = dBVg .... (ii)

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

From Eq. (ii) $${d_B} = {T \over {Vg}} + {d_F}$$. Hence dB > dF. 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

### IIT-JEE 2011 Paper 1 Offline

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.

## Explanation

Let T1, T2, T3 and T4 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 T2 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 $$d{Q_C}/dt = d{Q_B}/dt + d{Q_D}/dt$$.

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