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NEET NEET Questions — Free MCQ Practice

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A black metal foil is warmed by radiation from a small sphere at temperature T and at a distance d. It is found that the power received by the foil is `P'. If both the temperature and the distance are doubled, the power received by the foil will be

A.

16P

B.

4P

C.

2P

D.

P

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Explanation

(b) Energy received per second i.e., power

$P \propto (T^{4} - T_{0}^{4}) \Rightarrow P \propto T^{4} (∵ T_{0} < < T)$

Also energy received per sec (p) ∝ $\frac{1}{d^{2}}$

(inverse square law)

$\Rightarrow P \propto \frac{T^{4}}{d^{2}} \Rightarrow \frac{P_{1}}{P_{2}} = {(\frac{T_{1}}{T_{2}})}^{2} \times {(\frac{d_{2}}{d_{1}})}^{2} \Rightarrow \frac{P}{P_{2}} = {(\frac{T}{2 T})}^{2} \times {(\frac{2 d}{d})}^{2} = \frac{1}{4} \Rightarrow P_{2} = 4 P$

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Two metallic spheres $S_{1}$ and $S_{2}$ are made of the same material and have identical surface finish. The mass of $S_{1}$ is three times that of $S_{2}$ . Both the spheres are heated to the same high temperature and placed in the same room having lower temperature but are thermally insulated from each other. The ratio of the initial rate of cooling of $S_{1}$ to that of $S_{2}$ is

A.

$1 / 3$

B.

${(1 / 3)}^{1 / 3}$

C.

$1 / \sqrt{3}$

D.

$\sqrt{3} / 1$

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Explanation

(b) Rate of cooling (R) $= \frac{∆ θ}{t} = \frac{Aεσ (T^{4} - T_{0}^{4})}{mc}$

$\Rightarrow R \propto \frac{A}{m} \propto \frac{Area}{volume} \propto \frac{r^{2}}{r^{3}} \propto \frac{1}{r} \Rightarrow (R) \propto \frac{1}{r} \propto \frac{1}{m^{1 / 3}} [∵ m = ρ \times \frac{4}{3} {πr}^{3} \Rightarrow r \propto m^{1 / 3}] \Rightarrow \frac{R_{1}}{R_{2}} = {(\frac{m_{2}}{m_{1}})}^{1 / 3} = {(\frac{1}{3})}^{1 / 3}$

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Three discs A, B and C having radii 2m, 4m, and 6m respectively are coated with carbon black on their other surfaces. The wavelengths corresponding to maximum intensity are 300 nm, 400 nm and 500 nm, respectively. The power radiated by them are $Q_{a}$ , $Q_{b}$ , and $Q_{c}$ respectively

A.

$Q_{a}$ is maximum

B.

$Q_{b}$ is maximum

C.

$Q_{c}$ is maximum

D.

$Q_{a} = Q_{b} = Q_{c}$

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Explanation

(b) Radiated power $P = {AεσT}^{4} \Rightarrow P \propto {AT}^{4}$
From Wein’s law, $λ_{m} T = constant \Rightarrow T \propto \frac{1}{λ_{m}}$

$∴ P \propto \frac{A}{{(λ_{m})}^{4}} \propto \frac{r^{2}}{{(λ_{m})}^{4}} \Rightarrow Q_{A} : Q_{B} : Q_{C} = \frac{2^{2}}{{(300)}^{4}} : \frac{4^{2}}{{(400)}^{4}} : \frac{6^{2}}{{(500)}^{4}}$

$∴$ $Q_{B}$ will be maximum.

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A solid sphere and a hollow sphere of the same material and size are heated to the same temperature and allowed to cool in the same surroundings. If the temperature difference between each sphere and its surroundings is same , then

A.

The hollow sphere will cool at a faster rate for all values of T

B.

The solid sphere will cool at a faster rate for all values of T

C.

Both spheres will cool at the same rate for all values of T

D.

Both spheres will cool at the same rate only for small values of T

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Explanation

(a) Rate of cooling $\frac{∆ θ}{t} = \frac{Aεσ (T^{4} - T_{0}^{4})}{mc}$
As surface area, material and temperature difference are same, so rate of loss of heat is same in both the spheres. Now in this case rate of cooling depends on mass.

$\Rightarrow$ Rate of cooling $\frac{∆ θ}{t} \propto \frac{1}{m}$

$∵ m_{solid} > m_{hollow}$ . Hence hollow sphere will cool fast.

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A solid copper cube of edges 1 cm is suspended in an evacuated enclosure. Its temperature is found to fall from 100 $° C$ to 99 $° C$ in 100 s . Another solid copper cube of edges 2 cm, with similar surface nature, is suspended in a similar manner. The time required for this cube to cool from 100 $° C$ to 99 $° C$ will be approximately -

A.

25 s

B.

50 s

C.

200 s

D.

400 s

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Explanation

(c) Rate of cooling $\frac{∆ θ}{t} = \frac{Aeσ (T^{4} - T_{0}^{4})}{mc}$

$\Rightarrow t \propto \frac{m}{A} [∵ ∆ θ, t, σ (T^{4} - T_{0}^{4} are constant)]$

$\Rightarrow t \propto \frac{m}{A} \propto \frac{Volume}{Area} \propto \frac{a^{3}}{a^{2}} \Rightarrow t \propto a \Rightarrow \frac{t_{1}}{t_{2}} = \frac{a_{1}}{a_{2}} \Rightarrow \frac{100}{t_{2}} = \frac{1}{2} \Rightarrow t_{2} = 200 \sec$

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A body initially at 80 $° C$ cools to 64 $° C$ in 5 minutes and to 52 $° C$ in 10 minutes. The temperature of the body after 15 minutes will be

A.

42.7 $° C$

B.

35 $° C$

C.

47 $° C$

D.

40 $° C$

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Four identical rods of same material are joined end to end to form a square. If the temperature difference between the ends of a diagonal is 100 $° C$ , then the temperature difference between the ends of other diagonal will be -

A.

0 $° C$

B.

$\frac{100}{l} ° C$ ; where l is the length of each rod

C.

$\frac{100}{2 l} ° C$

D.

100 $° C$

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A cylindrical rod with one end in a steam chamber and the other end in ice results in melting of 0.1 gm of ice per second. If the rod is replaced by another with half the length and double the radius of the first and if the thermal conductivity of material of second rod is $\frac{1}{4}$ that of first, the rate at which ice melts in gm/sec will be -

A.

3.2

B.

1.6

C.

0.2

D.

0.1

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Explanation

(c) $\frac{Q}{t} = \frac{KA ∆ θ}{l} \Rightarrow \frac{mL}{t} = \frac{K ({πr}^{2}) ∆ θ}{l}$

$\Rightarrow$ Rate of melting of ice $(\frac{m}{t}) \propto \frac{{Kr}^{2}}{l}$

Since for second rod K becomes $\frac{1}{4} th$ , r becomes double and length becomes half, so rate of melting will be twice i.e.

${(\frac{m}{t})}_{2} = 2 {(\frac{m}{t})}_{1} = 2 \times 0.1 = 0.2 gm / \sec$

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One end of a copper rod of length 1.0 m and area of cross-section $10^{- 3} m^{2}$ is immersed in boiling water and the other end in ice. If the coefficient of thermal conductivity of copper is 92 cal/m-s $° C$ and the latent heat of ice is $8 \times 10^{4}$ cal/kg, then the amount of ice which will melt in one minute is -

A.

$9.2 \times 10^{- 3}$ kg

B.

$8 \times 10^{- 3}$

C.

$6.9 \times 10^{- 3}$

D.

$5.4 \times 10^{- 3}$

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Explanation

(c) Heat transferred in one minute is utilised in melting the ice so, $\frac{KA (θ_{1} - θ_{2}) t}{l} = m \times L$

$\Rightarrow m = \frac{10^{- 3} \times 92 \times (100 - 0) \times 60}{1 \times 8 \times 10^{4}} = 6.9 \times 10^{- 3} kg$

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An ice box used for keeping eatable cold has a total wall area of 1 ${metre}^{2}$ and a wall thickness of 5.0 cm. The thermal conductivity of the ice box is K = 0.01 joule/metre-s- $° C$ . It is filled with ice at 0 $° C$ along with eatables on a day when the temperature is 30°C. The latent heat of fusion of ice is $334 \times 10^{3} j o u l e s / k g$ . The amount of ice melted in one day is ( 1 day = 86,400 seconds )

A.

776 gms

B.

7760 gms

C.

11520 gms

D.

1552 gms

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Explanation

(d) $\frac{dQ}{dt} = \frac{KA}{l} dθ = \frac{0.01 \times 1}{0.05} \times 30 = 6 J / \sec$

Heat transferred in one day (86400 sec)

$θ = 6 \times 86400 = 518400 J$

Now Q = mL $\Rightarrow m = \frac{Q}{L} = \frac{518400}{334 \times 10^{3}} = 1.552 kg = 1552 g$

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