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NEET NEET Questions — Free MCQ Practice

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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

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

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A sink, that is the system where heat is rejected, is essential for the conversion of heat into work. From which law the above inference follows?

A.

zeroth

B.

First

C.

Second

D.

Third

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Explanation

$I I^{n d}$ law of thermodynamics

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An ideal gas with adiabatic exponent y is heated at constant pressure and it absorbs Q heat. What fraction of this heat is used to perform external work

A.

$γ$

B.

$\frac{γ}{γ}$

C.

$1 - \frac{1}{γ}$

D.

$γ - 1$

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Explanation

3.

$\begin{aligned} \frac{d U}{d Q} = \frac{1}{γ} \\ \frac{d Q - d W}{d Q} = \frac{1}{γ} \\ \frac{d W}{d Q} = (1 - \frac{1}{γ}) \end{aligned}$

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A Carnot engine working between 400K and 800K has a work output of 900J per cycle. The amount of heat energy supplied to engine from the source per cycle is

A.

900 J

B.

1800 J

C.

450 J

D.

2700 J

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Explanation

$(2) 1 - \frac{400}{800} = \frac{900}{Q_{1}}$

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Temperature is defined by

A.

First Law of thermodynamics

B.

Second Law of thermodynamics

C.

Third Law of thermodynamics

D.

Zeroth Law of thermodynamics

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Explanation

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If 32 gm of $O_{2}$ at $27 ° C$ is mixed with 64 gm of $O_{2}$ at $327 ° C$ in an adiabatic vessel, then the final temperature of the mixture will be :

A.

$200 ° C$

B.

$227 ° C$

C.

$314.5 ° C$

D.

$235.5 ° C$

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Explanation

$\frac{n_{1} f R T_{1}}{2} + \frac{n_{2} f R T_{2}}{2} = \frac{(n_{1} + n_{2}) f R T_{3}}{2} \frac{1 \times 300}{2} + \frac{2 \times 600}{2} = \frac{3 \times T}{2} T = 500 K (227 {}_{o}C)$

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