Three bodies A, B and C of masses m, m and sq | vedclass

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(D) Let the constant rate of heat supply be $\frac{dQ}{dt} = K$.From the principle of calorimetry,$dQ = ms \Delta 	heta$.Since $	heta_0 = 0$,we have

Vedclass · Accepted Answer

$S_B = S_C > S_A$

Vedclass · Answer

(D) Let the constant rate of heat supply be $\frac{dQ}{dt} = K$.From the principle of calorimetry,$dQ = ms \Delta 	heta$.Since $	heta_0 = 0$,we have $dQ = ms 	heta$.Integrating or considering the rate,$K t = ms 	heta$,which gives $	heta = \left( \frac{K}{ms} ight) t$.Comparing this with the equation of a line $y = mx$,the slope is $	ext{slope} = \frac{K}{ms} = 	an \alpha$,where $\alpha$ is the angle the graph makes with the time axis.Thus,$S = \frac{K}{m 	an \alpha}$.For body $A$ (mass $m$,angle $60^{\circ}$): $S_A = \frac{K}{m 	an 60^{\circ}} = \frac{K}{m \sqrt{3}}$.For body $B$ (mass $m$,angle $45^{\circ}$): $S_B = \frac{K}{m 	an 45^{\circ}} = \frac{K}{m}$.For body $C$ (mass $\sqrt{3}m$,angle $30^{\circ}$): $S_C = \frac{K}{(\sqrt{3}m) 	an 30^{\circ}} = \frac{K}{\sqrt{3}m 	imes \frac{1}{\sqrt{3}}} = \frac{K}{m}$.Comparing the values,we get $S_B = S_C = \frac{K}{m}$ and $S_A = \frac{K}{m \sqrt{3}}$.Since $\sqrt{3} > 1$,it follows that $\frac{K}{m \sqrt{3}} Therefore,$S_A  S_A$.

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