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6.System of Particles and Rotational Motion
medium
A disc of mass $3 \,kg$ rolls down an inclined plane of height $5 \,m$. The translational kinetic energy of the disc on reaching the bottom of the inclined plane is .......... $J$
A
$50$
B
$100$
C
$150$
D
$175$
Solution
(b)
Using mechanical energy conservation
$m g h=\frac{1}{2} m v^2+\frac{1}{2} l \omega^2$
$3(5)(10)=\frac{1}{2} m v^2+\frac{1}{2} m l^2\left(\frac{v^2}{l^2}\right)$
$150=\frac{3}{4} m v^2$
$m v^2=200$
$\frac{1}{2} m v^2=100 J = K \cdot E _{\cdot} \cdot \text { Translation }$
Standard 11
Physics
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