InterviewSolution

Explore topic-wise InterviewSolutions in Class 11.

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your Class 11 knowledge and support exam preparation. Choose a topic below to get started.

1.	Three rods of material `x` and three of material `y` are connected as shown in figure. All the rods are identical in length and cross sectional area. If the end `A` is maintained at `60^(@)C` and the junction `E` at `10^(@)C` , calculate the temperature of the junction `B`. The thermal conductivity of `x` is `800Wm^(-1).^(@)C^(-1)` and that of `y` is `400Wm^(-1).^(@)C^(-1)` .
Answer» Correct Answer - B::C::D<br>Thermal resistance, `R = l/(KA)`<br> `R_x/R_y = K_y/ K_x ` (as `l_x = l_y` and `A_x = A_y`)<br> `=0.46/0.92`<br> `=1/2`<br> So, if `R_x = R` then` R_y = 2R`<br> CEDB forms a balanced wheatstone bridge, i.e. `T_C - T_D` and no heat flows through CD.<br> `:. 1/R_(BE) = 1/(R+R) + 1/(2R + 2R)`<br> or `R_(BE) = 4/3 R`<br> The total resistance between A and E will be<br> `R_(AE) = R_(AB) + R_(BE) = 2R + 4/3R = 10/3R`<br> `:.` Heat current between A and E is<br> `H = ((DeltaT)_(AE))/R_(AE) = (60-10)/((10//3)R) = 15/R`<br> Now, if `T_B` is the temperature at B,<br> `H_(AB) = ((DeltaT)_(AB))/R_(AB)`<br> or `15/R = (60 - T_(B))/(2R)`<br> or `T_(B) = 30^@C`<br> Further,<br> `H_(AB) = H_(BC) + H_(BD)`<br> or `15/R = (30-T_(C))/R + (30-T_(D))/(2R) [T_C = T_D = T(say)]`<br> or `15 = (30-T) + (30-T)/2`<br> Solving this, we get `T=20^@C`<br> or `T_C = T_D = 20^@C`

2.	Three rods of identical cross-sectional area and made from the same metal form the sides of an isosceles triangle ABC right angled at B as shown in the figure. The points A and B are maintained at temperature T and `(sqrt2)T` respectively in the steady state. Assuming that only heat conduction takes place, temperature of point C will be A. `T/((sqrt2)-1)`B. `T/((sqrt2)+1)`C. `(3T)/((sqrt 2)+1)`D. `T/((sqrt3)(sqrt2 -1))`
Answer» Correct Answer - C<br><br> `H_1 = H_2`<br> `:. (sqrt(2)T -T_C)/((l//KA)) = (T_C-T)/((sqrt(2)l//KA)`<br> Solving we get,<br> `T_C = (3T)/((sqrt2) +1)`.

3.	Four identical rods `AB, CD, CF and DE` are joined as shown in figure. The length, cross-sectional area and thermal conductivity of each rod are `l, A and K` respectively. The ends `A, E and F` are maintained at temperature `T_(1)` , `T_(2)` and `T_(3)` respectively. Assuming no loss of heat to the atmosphere, find the temperature at `B`.
Answer» Correct Answer - A::B::C<br><br> `H_1 + H_2 = H_3`<br> `((T_3-T)/(3R//2)) + ((T_2-T)/(3R//2)) = (T-T_1)/R`<br> `:. T = ((3T_1+2(T_2+T_3))/7)` .

4.	A long rod has one end at `0^@C` and other end at a high temperature. The coefficient of thermal conductivity varies with distance from the low temperature end as `K = K_0(1+ax)`, where `K_0 = 10^2` SI unit and `a = 1m^-1` . At what distance from the first end the temperature will be `100^@C`? The area of cross-section is `1cm^2` and rate of heat conduction is 1 W.A. 2.7 mB. 1.7mC. 3 mD. 1.5 m
Answer» Correct Answer - B<br>`H = (TD)/R`<br> `:. R = (TD)/H = (100 -0)/1`<br> `=100kW^-1`<br> Now, `R= int_(0)^x dR = int_(0)^x ((dx)/(K_0(1+ax)A))`<br> or `100 = int_(0)^x (dx)/(10^2(1+x)(10^-4))`<br> Solving this equation we get,<br> `x = 1.7 m.`

5.	An ice cube of mass 0.1 kg at `0^@C` is placed in an isolated container which is at `227^@C`. The specific heat s of the container varies with temperature T according to the empirical relation `s=A+BT`, where `A= 100 cal//kg-K and B = 2xx (10^-2) cal//kg-K^2`. If the final temperature of the container is `27^@C`, determine the mass of the container. (Latent heat of fusion for water = `8xx (10^4) cal//kg`, specific heat of water `=10^3 cal//kg-K`).
Answer» Correct Answer - D<br>Let m be the mass of the container.<br> Initial temperature of container,<br> `T_i = (227 + 273) = 500 K`<br> and final temperature of container,<br> `T_f = (27 + 273) = 300 K `<br> Now, heat gained by the ice cube = heat lost by the container<br> `:. (0.1)(8xx (10^4)) + (0.1)(10^3)(27) = -m (int_(500)^300)(A+BT)dT`<br> or `10700 = -m [AT + (BT^2)/2]_(500)^(300)`<br> After substituting the values of A and B and the proper limits, we get<br> `m=0.495 kg` .

6.	1 g of ice at `0^@C` is mixed with 1 g of steam at `100^@C`. After thermal equilibrium is achieved, the temperature of the mixture isA. `100^@C`B. `55^@C`C. `75^@C`D. `0^@C`
Answer» Correct Answer - A<br>Heat taken by 1 g ice in transformation from ice at `0^@C` to water at` 100^@C` is<br> `Q = mL + msDeltatheta`<br> `= (1)(80) + (1)(1)(100)`<br> = 180 cal<br> Mass of steam condensed to give this much heat is<br> `m = Q/L = 180/540 = 1/3g`<br> This is less than 1g or total mass of steam.<br> Therefore, whole steam is not condensed and<br> mixture temperature is `100^@C`.

7.	Two identical conducting rods are first connected independently to two vessels, one containing water at `100^@C` and the other containing ice at `0^@C`. In the second case, the rods are joined end to end and connected to the same vessels. Let `q_1 and q_2` gram per second be the rate of melting of ice in the two cases respectively. The ratio `q_1/q_2` is (a) `1/2` (b)`2/1` (c)`4/1` (d)`1/4`
Answer» Correct Answer - A<br>(c) `(dQ)/(dt) = L(dm)/(dt) or ("Temperature difference")/("Thermal resistance") = L ((dm)/(dt))`<br> or `(dm)/(dt) prop 1/("Thermal resistance" ) rArr q prop 1/R`<br> In the first case rods are in parallel and thermal resistance is `R/2` while in second case rods are in series and thermal resistance is 2R.<br> `(q_1/q_2) = (2R)/(R//2) = 4/1`

8.	The specific heat of a metal at low temperatures varies according to `S = aT^3`, where a is a constant and T is absolute temperature. The heat energy needed to raise unit mass of the metal from temperature `T = 1 K` to `T = 2K` isA. `3a`B. `(15a)/4`C. `(2a)/3`D. `(13a)/4`
Answer» Correct Answer - B<br>`Q = int (dQ) = int_(1)^2 msdT`<br> `=int_(1)^2 (1)(aT)^3 dT = (15a)/4`

9.	Assertion : Specific heat of any substance remains constant at all temperatures. Reason : It is given by ` s= 1/m * (dQ)/(dt)` .A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true.
Answer» Correct Answer - D<br>Specific heat is a function of temperature. But for most of the substances variation with temperature is almost negligible.

10.	Two liquids are at temperatures `20^@C and 40^@C`. When same mass of both of them is mixed, the temperature of the mixture is `32^@C`. What is the ratio of their specific heats?A. `1//3`B. `2//5`C. `3//2`D. `2//3`
Answer» Correct Answer - D<br>`Q_1 = Q_2`<br> `:. ms_1 (32 -20) = ms_2 (40 - 32)`<br> `s_1/s_2 = 8/12 = 2/3. `

11.	Two conducting rods when connected between two points at constant but different temperatures separately, the rate of heat flow through them is `q_1 and q_2`.A. When they are connected in series, the net rate of heat flow is `q_1 + q_2`B. When they are connected in series, the net rate of heat flow is `((q_1q_2)/(q_1+q_2))`C. When they are connected in parallel, the net rate of heat flow is `q_1 + q_2`D. When they are connected in parallel, the net rate of heat flow is `((q_1q_2)/(q_1+q_2))`
Answer» Correct Answer - B::C<br>`q_1 = (TD)/R_1`<br> `R_1 = (TD)/q_1`<br> Similarly, `R_2 = (TD)/q_2`<br> In series,<br> `q_s = (TD)/(R_1+R_2)= (TD)/(((TD)/q_1) + ((TD)/q_2))`<br> `=(q_1q_2)/(q_1+q_2)`<br> In parallel, `q_p = (TD)/R_(net) = (TD) (1/R_(net))`<br> `=TD (1/R_1 + 1/R_2)`<br> `=TD ( q_1/(TD) + q_2/(TD)) = q_1 + q_2` .