JEE Physics Transmission Of Heat MCQs Set 02

Answer: (b) \( 10^{7} \) K

Explanation:
1. \( T = \frac{2.93\times10^{-3}}{2.93\times10^{-10}} = 10^7 \) K.

Teacher's Note:
1. Divide Wien's constant by the peak wavelength.
2. The very short wavelength implies a very high temperature.

Answer: (a) \( (T_2/T_1)^2 \)

Explanation:
1. Power \( = \sigma\,4\pi r^2T^4 \). Equal power gives \( r_1^2T_1^4 = r_2^2T_2^4 \).
2. So \( r_1/r_2 = (T_2/T_1)^2 \).

Teacher's Note:
1. Use the Stefan-Boltzmann law \( P = \sigma A T^4 \).
2. Take the square root of the ratio of \( T^4 \) terms.

Answer: (a) 1 : 1

Explanation:
1. Power \( \propto r^2T^4 \). First: \( 16\times2000^4 \); second: \( 1\times4000^4 = 16\times2000^4 \).
2. The powers are equal, so the ratio is 1 : 1.

Teacher's Note:
1. Compare \( r^2T^4 \) for each body.
2. Doubling the temperature gives a factor 16 that offsets a radius ratio of 4.

Answer: (a) The sphere

Explanation:
1. For the same mass, the sphere has the smallest surface area, so it loses heat at the slowest rate.
2. It therefore takes the longest to cool.

Teacher's Note:
1. Rate of cooling grows with exposed surface area.
2. The sphere is the shape with the least surface area for a given volume.

Answer: (c) \( T_1\lt T_2\lt T_3 \)

Explanation:
1. By Newton's law, the rate of cooling falls as the excess temperature over the surroundings falls.
2. Each equal 5 degree drop therefore takes longer than the one before: \( T_1\lt T_2\lt T_3 \).

Teacher's Note:
1. Cooling is fastest when the body is much hotter than its surroundings.
2. Equal temperature drops take longer as the body approaches room temperature.

Answer: (b) Equal to T minutes

Explanation:
1. Newton's law: drop \( \propto (\bar T-T_0)\,t \).
2. First case: \( 1 = k(61.5-30)T = 31.5\,kT \).
3. Second: \( 0.5 = k(45.75-30)t = 15.75\,kt \), so \( t = \frac{0.5}{15.75k} = \frac{1}{31.5k} = T \).

Teacher's Note:
1. Half the temperature drop at half the excess temperature takes the same time.
2. Use the mean temperature of the interval for the excess.

Answer: (b) \( 10^\circ \)C

Explanation:
1. Newton's law: \( \frac{10}{10} = k(55-T_0) \) and \( \frac{8}{10} = k(46-T_0) \).
2. Dividing: \( \frac{55-T_0}{46-T_0} = 1.25 \), so \( 55-T_0 = 57.5-1.25T_0 \) and \( T_0 = 10^\circ \)C.

Teacher's Note:
1. Apply Newton's law of cooling to each 10 minute interval using the mean temperature.
2. The ratio of the two equations removes the constant k.

Answer: (d) 1552 g

Explanation:
1. Heat entering \( Q = \frac{KA\Delta T\,t}{d} = \frac{0.01\times1\times30\times86400}{0.05} = 5.184\times10^5 \) J.
2. Mass melted \( = \frac{5.184\times10^5}{3.34\times10^5} = 1.552 \) kg \( = 1552 \) g.

Teacher's Note:
1. Total heat conducted equals power times time.
2. Divide by the latent heat to get the mass of ice.

Answer: (a) (i) and (ii) only

Explanation:
1. Equal power: \( e_AT_A^4 = e_BT_B^4 \), so \( T_B = 5802\times(0.01/0.81)^{1/4} = 5802/3 = 1934 \) K.
2. \( \lambda_A = \frac{2.898\times10^{-3}}{5802} = 0.5\ \mu\)m and \( \lambda_B = \frac{2.898\times10^{-3}}{1934} = 1.5\ \mu\)m, a shift of 1.0 \(\mu\)m.
3. Statements (i) and (ii) are correct.

Teacher's Note:
1. Combine the Stefan-Boltzmann and Wien laws.
2. Check the 1 micrometre shift to confirm the data are consistent.

Answer: (d) (ii) and (iv)

Explanation:
1. The strips of thickness t differ in extension by \( |\alpha_B-\alpha_C|\Delta T\,L \), and this difference over the separation gives the curvature.
2. \( R \approx \frac{t}{(\alpha_B-\alpha_C)\Delta T} \), which is inversely proportional to both \( \Delta T \) and \( |\alpha_B-\alpha_C| \).

Teacher's Note:
1. Greater temperature rise or greater mismatch gives a tighter curve (smaller R).
2. This is the principle of a bimetallic thermostat.

Answer: (c) A is true, but R is false.

Explanation:
1. A small hole in a uniformly heated cavity emits black body radiation, so the assertion is true.
2. Metals are not perfect black bodies (they reflect), so the reason is false.

Teacher's Note:
1. A cavity with a small opening approximates an ideal black body.
2. Real metal surfaces reflect a lot of radiation.

Answer: (c) A is true, but R is false.

Explanation:
1. Stefan's law, \( E\propto T^4 \), applies to the sun to a good approximation, so A is true.
2. The sun radiates very nearly like a black body, so R is false.

Teacher's Note:
1. The sun is a close approximation to an ideal black body.
2. Stefan's law is exact only for black bodies.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A hotter star peaks at a shorter wavelength, and blue is a shorter wavelength than red.
2. Wien's law \( \lambda_mT = b \) is exactly this reasoning, so R explains A.

Teacher's Note:
1. Peak wavelength shifts to the blue as temperature rises.
2. Star colour is a quick temperature indicator.

Answer: (c) A is true, but R is false.

Explanation:
1. Water has a very high specific heat, so it warms and cools slowly, which moderates coastal temperatures. A is true.
2. Water is a poor conductor of heat, so R is false.

Teacher's Note:
1. The reason is high specific heat, not high conductivity.
2. Land heats and cools much faster than water.

Answer: (c) A is true, but R is false.

Explanation:
1. Hot air rises by convection, so the region above the fire is hotter. A is true.
2. Air is a poor conductor, so the transfer is by convection and not conduction. R is false.

Teacher's Note:
1. Convection currents carry heat upwards.
2. Air is a very poor conductor of heat.

Answer: (a) \( \frac{2}{3} \)

Explanation:
1. Wien's law: \( T\propto 1/\lambda_m \).
2. \( \frac{T_{sun}}{T_{star}} = \frac{360}{540} = \frac{2}{3} \).

Teacher's Note:
1. The star with the shorter peak wavelength is the hotter one.
2. Invert the wavelength ratio to get the temperature ratio.