Read the following statements:A. When the sma | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) $A.$ According to Newton's law of cooling,the rate of loss of heat $\frac{dQ}{dt} \propto \Delta T$. If $\Delta T$ is doubled,the rate of loss of

Vedclass · Accepted Answer

$A, B, C$ only

Vedclass · Answer

(A) $A.$ According to Newton's law of cooling,the rate of loss of heat $\frac{dQ}{dt} \propto \Delta T$. If $\Delta T$ is doubled,the rate of loss of heat becomes twice. Thus,$A$ is correct.$B.$ The power radiated by a body is $P = \sigma e A T^4$. The ratio of radiation emitted is $\frac{H_P}{H_Q} = \left(\frac{T_P}{T_Q}\right)^4$. Converting temperatures to Kelvin: $T_P = 10 + 273 = 283 K$ and $T_Q = 20 + 273 = 293 K$. Thus,$\frac{H_P}{H_Q} = \left(\frac{283}{293}\right)^4 \approx (0.9658)^4 \approx 0.87$. However,using the approximation $\frac{293}{283} \approx 1.035$,then $(1.035)^4 \approx 1.15$. Thus,$1:1.15$ is correct. $B$ is correct.$C.$ Efficiency $\eta = 1 - \frac{T_L}{T_H} = 1 - \frac{100}{400} = 1 - 0.25 = 0.75$ or $75\%$. Thus,$C$ is correct.$D.$ Since $\frac{dQ}{dt} \propto \Delta T$,if $\Delta T$ is quadrupled,the rate of loss of heat becomes four times,not twice. Thus,$D$ is incorrect.Therefore,statements $A, B,$ and $C$ are correct.

Column $I$	Column $II$
$(A)$ An insulated container has two chambers separated by a valve. Chamber $I$ contains an ideal gas and Chamber $II$ has a vacuum. The valve is opened.	$(p)$ The temperature of the gas decreases
$(B)$ An ideal monoatomic gas expands to twice its original volume such that its pressure $P \propto V^{-2}$	$(q)$ The temperature of the gas increases or remains constant
$(C)$ An ideal monoatomic gas expands to twice its original volume such that its pressure $P \propto V^{-4/3}$	$(r)$ The gas loses heat
$(D)$ An ideal monoatomic gas expands such that its pressure $P$ and volume $V$ follow the behavior shown in the graph	$(s)$ The gas gains heat

Similar Questions

$A$ gas expands with temperature according to the relation,$V=k T^{2/3}$,where $k$ is a constant. Work done when the temperature changes by $60 \ K$ is ($R=$ universal gas constant.) (in $R$)

An ideal gas is taken from state $1$ to state $2$ through optional paths $A, B, C$ and $D$ as shown in the $P-V$ diagram. Let $Q, W$ and $\Delta U$ represent the heat supplied,work done,and change in internal energy of the gas,respectively. Then:

Initial pressure and volume of a gas are $P$ and $V$ respectively. First it is expanded isothermally to volume $4V$ and then compressed adiabatically to volume $V$. The final pressure of the gas will be (given $\gamma = 3/2$): (in $,P$)

Heat is supplied to a diatomic gas at constant pressure. The ratio of $\Delta Q : \Delta U : \Delta W$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module