\[(a)\;k_1=k_2 \quad (b)\;k_1 < k_2 \quad (c)\; k_1 > k_2 \quad (d)\;data\;insufficient\]

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Work done against friction

$W=\mu mg \cos \theta\;s$

$W=\mu mg \cos \theta\;\large\frac{h}{\sin \theta}$

$W=\mu mgh\; \cot \theta$

$\cot \theta_1=\cot 30^{\circ}=\sqrt 3$

$\cot \theta_2=\cot 60^{\circ}=\large\frac{1}{\sqrt 3}$

$W_1 > W_2$

Kinetic energy $k=mgh-W$

$\therefore $ kinetic energy in 1st case $(\theta=30^{\circ})$ will be less than kinetic energy when $(\theta=60^{\circ})$

$k_1 < k_2$

Hence b is the correct answer.

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