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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. |
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`).A. `0.495 kg`B. `0.595 kg`C. `0.695 kg`D. `0.795 kg` |
Answer» Correct Answer - A<br>Heat lost by container `= iint_(500)^(300) m_(C)(A + BT) dT`<br> `= - m_(C) [At + (BT^(2))/(2)]_(500)^(300) = 21600 m_(C)`<br> Heat gained by ice `= m_(ice) L + m_(ice)s_("water")DeltaT`<br> `= 0.1 xx 8 xx 10^(4) + 0.1 xx 10^(3) xx 27`<br> `= 10700 cal`<br> According to principal of calorimetry<br> Heat lost by container = Heat gained by ice<br> `21600 m_(C) = 10700`<br> or `m _(C) = 0.495` kg | |
2. |
Which of the following circular rods (given radius `r` and length `l`) each made of the same material and whose ends are maintained at the same temperature will conduct most heat?A. `r = 2r_(0), l = 2l_(0)`B. `r = 2r_(0), l = l_(0)`C. `r = r_(0), l = l_(0)`D. `r = r_(0), l = 2l_(0)` |
Answer» Correct Answer - B<br>Heat conduction through a rod is<br> `H = (Delta Q)/(Deltat) = KA ((T_1-T_(2))/(l))`<br> `rArr H prop (r^(2))/(l)` ..(i)<br> When `r = 2r_(0), l = 2l_(0)`<br> `H prop ((2r_(0))^(2))/(2l_(0)) rArr H prop (2r_(0)^(2))/(l_0)`<br> (b) when `r = 2r_(0), l = l_(0)`<br> `H prop ((2r_(0))^(2))/(l_(0)) rArr H prop (4r_(0)^(2))/(l_0)`<br> (c) When `r = r_(0), l = l_(0)`<br> `H prop ((r_(0))^(2))/(l_(0))`<br> (d) When `r = r_(0), l = 2l_(0)`<br> `H prop ((r_(0))^(2))/(2l_(0))`<br> It is obvisous heat conduction will be more in case (b) | |
3. |
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`).A. `0.495 kg`B. `0.224 kg`C. `0.336 kg`D. `0.621 kg` |
Answer» Correct Answer - A<br>Heat recived by ice is<br> `Q_(1)=mL+mCDelta T = 10700cal`<br> Heat lost be the container is<br> `Q_(2) = underset(300)overset(500)(int) m_(C) (A+BT)dT = m_(C)[AT +(BT^2)/(2)]_(300)^(500)`<br> `=21600m_(C)`<br> By principle of calorimetry, `Q_(1)=Q_(2)`<br> `rArr m_(c) = 0.495 kg`. | |
4. |
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`A. `1/2`B. `2/1`C. `4/1`D. `1/4` |
Answer» Correct Answer - C<br>Initially the rods are placed in vessels as shown below<br> <br> `(Q)/(t) = ((theta_(1)-theta_(2)))/(R) rArr (Q/t)_(1) = (mL)/(t) = q_(1)L=((100-0))/((R)/(2))` ..(i)<br> Finally when rods are joined end to as shown<br> <br> `rArr (Q/t)_(2) = (mL)/(t) = q_(2)L = ((100-0))/(2R)` ..(ii)<br> Form equation (i) and (ii) , `(q_1)/(q_2) = 4/1`. | |
5. |
A cylinder of radius R made of a material of thermal conductivity `K_1` is surrounded by a cylindrical shell of inner radius R and outer radius 2R made of a material of thermal conductivity `K_2`. The two ends of the combined system are maintained at two different temperatures. There is no loss of heat across the cylindrical surface and the system is in steady state. The effective thermal conductivity of the system isA. `K_(1) + K_(2)`B. `(K_(1) + 3K_(2))/(4)`C. `(K_(1) K_(2))/(K_(1) + K_(2))`D. `(3K_(1) + K_(2))/(4)` |
Answer» Correct Answer - B<br>`(dQ)/(dt) = (dQ_(1))/(dt) + (dQ_(2))/(dt)`<br> `K(4 pi R^(2)(theta_(1)-theta_(2)))/(L) = K_(1)(pi R^(2)(theta_(1)-theta_(2)))/(L)`<br> ` K_(3)(3pi R^(2)(theta_(1)-theta_(2)))/(L)`<br> or , `4K=K_(1)+3K_(2)`<br> or, `K=(K_(1)+3K_(2))//4` . | |