1.

A steel ball of mass m_(1)=1kg moving with velocity 50ms^(-1) collides with another ball of mass m_(2)=200g lying on the ground. Due the collision, the KE is lost and their internal energies change equally and T_(1) and T_(2) are the temperature changes of masses m_(1) and m_(2) respectively. If the specific heat of steel is unity and J="4.18 J cal"^(-1), then

Answer»

`T_(1)=7.1^(@)C and T_(2)=1.47^(@)C`
`T_(1)=1.47^(@)C and T_(2)=7.1^(@)C`
`T_(1)=3.4^(@)C and T_(2)=17.0^(@)C`
`T_(1)=17.0^(@)C and T_(2)=3.4^(@)C`

ANSWER :C


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