A) \[\frac{Q}{4}\]
B) \[\frac{Q}{16}\]
C) 2Q
D) \[\frac{Q}{2}\]
Correct Answer: B
Solution :
The rate of heat flow is given by \[\frac{Q}{t}=K.A.\,\frac{\Delta T}{\ell }\] Area of Original rod \[A=\pi \,{{R}^{2}}\] Areal of new rod\[A'=\frac{\pi \,{{R}^{2}}}{4}\]. 4 Volume of original rod will be equal to the volume of new rod. \[\therefore \,\,\,\pi {{R}^{2}}\ell =\pi {{\left( \frac{R}{2} \right)}^{2}}\,\ell '\] \[\Rightarrow \,\,\,\frac{\ell '}{\ell }=\frac{{{R}^{2}}}{\left( \frac{{{R}^{2}}}{4} \right)}=4\] \[\therefore \,\,\,\,\,\frac{Q'}{Q}=\frac{A'}{A}\,\,\frac{\ell }{\ell '}=\frac{1}{4}.\frac{1}{4}=\frac{1}{16}\] \[\therefore \,\,\,\,\,Q'=\frac{Q}{16}\]
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