A solid sphere of radius $R$ and a hollow sphere of inner radius $r$ and outer radius $R$ made of copper are heated to the same temperature and are allowed to cool in the same environment. Then, choose the $CORRECT$ statement
A solid sphere of radius $R$ and a hollow sphere of inner radius $r$ and outer radius $R$ made of copper are heated to the same temperature and are allowed to cool in the same environment. Then, choose the $CORRECT$ statement
- A
Hollow sphere cools faster
- B
Solid sphere cools faster
- C
Both the spheres attain room temperature at the same time
- D
The rate of loss of heat of the solid sphere is twice that of the hollow sphere
Similar Questions
A hollow copper sphere $S$ and a hollow copper cube $ C$ , both of negligible thin walls of same area, are filled with water at $90°C$ and allowed to cool in the same environment. The graph that correctly represents their cooling is
A hollow copper sphere $S$ and a hollow copper cube $ C$ , both of negligible thin walls of same area, are filled with water at $90°C$ and allowed to cool in the same environment. The graph that correctly represents their cooling is
A cup of tea cools from ${80^0}C$ to ${60^o}C$ in one minute. The ambient temperature is ${30^o}C$. In cooling from ${60^o}C$ to ${50^o}C$ it will take ....... $\sec$
A cup of tea cools from ${80^0}C$ to ${60^o}C$ in one minute. The ambient temperature is ${30^o}C$. In cooling from ${60^o}C$ to ${50^o}C$ it will take ....... $\sec$
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- [JEE MAIN 2020]
A cup of tea cools from $80\,^oC$ to $60\,^oC$ in one minute. The ambient temperature is $30\,^oC$. In cooling from $60\,^oC$ to $50\,^oC$, it will take ....... $\sec$
A cup of tea cools from $80\,^oC$ to $60\,^oC$ in one minute. The ambient temperature is $30\,^oC$. In cooling from $60\,^oC$ to $50\,^oC$, it will take ....... $\sec$
For a system with newtons law of cooling applicable the initial rate of cooling is $R^0\ C/sec$ find the time when temperature diff. $\Delta T_0 =$ initial temperature difference, is reduced to half.
For a system with newtons law of cooling applicable the initial rate of cooling is $R^0\ C/sec$ find the time when temperature diff. $\Delta T_0 =$ initial temperature difference, is reduced to half.