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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. |
A solid copper sphere (density rho and specific heat c) of radius r at an initial temperature `200K` is suspended inside a chamber whose walls are at almost `0K.` The time required for the temperature of the sphere to drop to 100K is ……. |
Answer» The energy emitted per second when the temperature of the copper sphere is `T` and the surrounding is temperature `T_(0)`<br> `=sigma(T^(4)-T_(0)^(4))xxA` (i)<br> Here `T_(0)=0K`<br> We know that<br> `dQ= -mcdt`<br> `:. (dQ)/(dt)=-mc(dT)/(dt)` (ii)<br> Here the `-ve` sign shows that the temperature is decreasing with time.<br> Energy emitted per second from Eqs. (i) and (ii)<br> `=sigmaT^(4)A=-mc(dT)/(dt)`<br> `implies dta=-(mcdT)/(sigmaT^(4)A)=-(pxx(4)/(3)pir^(3)cdT)/(sigmaT^(4)xx4pir^(2))`<br> `[:. m=pxx(4)/(3)pir^(3)]`<br> `implies dt=-(prc)/(3sigma)(dT)/(T^(4))`<br> Integrating both sides,<br> `int_(0)^(t)dt=-(prc)/(3sigma)int_(200)^(100)(dT)/(T^(4))=-(prc)/(3sigma)[-(1)/(3T^(3))]_(200)^(100)`<br> `t=(prc)/(9sigma)[(1)/((100)^(3))-(1)/((200)^(3))]`<br> `t=(7prc)/((72xx10^(6)sigma))` | |
2. |
A monoatomic ideal gas, initially at temperature `T_1,` is enclosed in a cylinder fitted with a friction less piston. The gas is allowed to expand adiabatically to a temperature `T_2` by releasing the piston suddenly. If `L_1 and L_2` are the length of the gas column before expansion respectively, then `(T_1)/(T_2)` is given byA. `(L_1/L_2)^(2//3)`B. `L_1/L_2`C. `L_2/L_1`D. `(L_2/L_1)^(2//3)` |
Answer» Correct Answer - D<br>Here `TV^(gamma-1)=`constant<br> As `gamma=5//3` , hence `TV^(2//3)=`constant<br> Now `T_(1)L_(1)^(2//3)=T_(2)L_(2)^(2//3)` , `(because V prop L)`<br> Hence `(T_(1))/(T_(2))=((L_(2))/(L_(1)))^(2//3)` | |
3. |
A black body of temperature T is inside chamber of `T_0` temperature initially. Sun rays are allowed to fall from a hole in the top of chamber. If the temperature of black body (T) and chamber `(T_0)` remains constant, then A. Black body will absorb radiation.B. Black body will absorb less radiation.C. Black body will emit more energy.D. Black body will emit energy equal to energy absorbed by it. |
Answer» Correct Answer - A::D<br>Since the sun rays fall on the black body, it will absorb radiations and since its temperature is constant, it will emit radiations. The temperature is will remain same only when energy emitted is equal to energy absorbed. | |
4. |
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_1K_2)/((K_1+K_2))`C. `((K_1+3K_2))/(4)`D. `((3K_1+K_2))/(4)` |
Answer» Correct Answer - C<br>Let `R_(1)` and `R_(2)` be the thermal resistances of inner and outer portions. Since temperature difference at both ends is same, the resistances are in parallel. Hence,<br> <br> `(1)/(R)=(1)/(R_(1))+(1)/(R_(2))`<br> `(K(4piR^(2)))/(l)=(K_(1)(piR^(2)))/(l)+(K_(2)(3piR^(2)))/(l)rArr =(3K_(2)+K_(1))/(4)`. | |