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Physics NEET Questions — Free MCQ Practice

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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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A hot metallic sphere of radius r radiates heat. It's rate of cooling is

A.

Independent of r

B.

Proportional to r

C.

Proportional to $r^{2}$

D.

Proportional to 1/r

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Explanation

(d) Rate of cooling $R_{C} = \frac{dθ}{dt} = \frac{Aεσ (T^{4} - T_{0}^{4})}{mc}$

$\Rightarrow \frac{dθ}{dt} \propto \frac{A}{V} \propto \frac{r^{2}}{r^{3}} \Rightarrow \frac{dθ}{dt} \propto \frac{1}{r}$

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A solid copper sphere (density $ρ$ and specific heat capacity c) of radius r at an initial temperature 200K is suspended inside a chamber whose walls are at almost 0K. The time required (in $μ$ s) for the temperature of the sphere to drop to 100 K is

A.

$\frac{72}{7} \frac{rρc}{σ}$

B.

$\frac{7}{72} \frac{rρc}{σ}$

C.

$\frac{27}{7} \frac{rρc}{σ}$

D.

$\frac{7}{27} \frac{rρc}{σ}$

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Explanation

(b) $\frac{dT}{dt} = \frac{σA}{mcJ} (T^{4} - T_{0}^{4})$ [In the given problem fall in temperature of body dT=(200-100)=100 K, temp. of surrounding $T_{0}$ = 0K, Initial temperature of body T = 200 K]

$\frac{100}{dt} = \frac{σ 4 {πr}^{2}}{\frac{4}{3} {πr}^{3} ρcJ} (200^{4} - 0^{4}) \Rightarrow dt = \frac{rρcJ}{48 σ} \times 10^{- 6} s = \frac{rρc}{σ} . \frac{4.2}{48} \times 10^{- 6} = \frac{7}{80} \frac{rρc}{σ} μs â‰ƒ \frac{7}{72} \frac{rρc}{σ} μs [As J = 4.2]$

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A sphere and a cube of same material and same volume are heated upto same temperature and allowed to cool in the same surroundings. The ratio of the amounts of radiations emitted will be

A.

1 : 1

B.

$\frac{4 π}{3} : 1$

C.

${(\frac{π}{6})}^{1 / 3} : 1$

D.

$\frac{1}{2} {(\frac{4 π}{3})}^{\frac{2}{3}} : 1$

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Explanation

(c) $Q = σAt (T^{4} - T_{0}^{4})$

If T, $T_{0}$ , $σ$ and t are same for both bodies then

$\frac{Q_{sphere}}{Q_{cube}} = \frac{A_{sphere}}{A_{cube}} = \frac{4 {πr}^{2}}{6 a^{2}}$ ....(i)

But according to problem, volume of sphere = Volume of cube $\Rightarrow \frac{4}{3} {πr}^{3} = a^{3} \Rightarrow a = {(\frac{4}{3} π)}^{1 / 3} r$

Substituting the value of a in equation (i) we get

$\frac{Q_{sphere}}{Q_{cube}} = \frac{4 {πr}^{2}}{6 a^{2}} = \frac{4 {πr}^{2}}{6 {\{{(\frac{4}{3} π)}^{1 / 3} r\}}^{2}} = \frac{4 {πr}^{2}}{6 {(\frac{4}{3} π)}^{2 / 3} r^{2}} = {(\frac{π}{6})}^{1 / 3} : 1$

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A system is taken from state A to state B along two different paths 1 and 2. If the heat absorbed and work done by the system along these two paths are $Q_{1}, Q_{2} a n d W_{1}, W_{2}$ respectively, then

A.

$Q_{1} = Q_{2}$

B.

$W_{1} = W_{2}$

C.

$Q_{1} - W_{1} = Q_{2} - W_{2}$

D.

$Q_{1} + W_{1} = Q_{2} + W_{2}$

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Explanation

Internal energy be state function i.e. not depend the paths. From first law of thermodynamics, Q=U+W

so, $Q_{1} - W = Q_{2} - W$

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The ratio of the relative rise in pressure for adiabatic compression to that for isothermal compression is

A.

$γ$

B.

$\frac{1}{γ}$

C.

1- $γ$

D.

$\frac{1}{1 - γ}$

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Explanation

$PV = k . . . . . . . . Isothermal Process VdP + pdV = 0 {(\frac{dP}{P})}_{Isothermal} = \frac{- dV}{V} {PV}^{γ} = k V^{γ} dP + {γPV}^{γ - 1} dV = 0 . . . . . . . . Adiabatic process {(\frac{dP}{P})}_{Adiabatic} = \frac{- γdV}{V} \frac{{(\frac{dP}{P})}_{Adiabatic}}{{(\frac{dP}{P})}_{Isothermal}} = γ$

Physics MCQs for NEET — Practice Questions with Answers

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