A) \[h=\frac{(x+5)}{2}r\]
B) \[h=\frac{x}{2}r\]
C) \[h=r\]
D) \[h=\left( \frac{x+1}{2} \right)r\]
Correct Answer: A
Solution :
[a] KE of Blocks at \[B=PE\] at A-PE at B \[\frac{1}{2}m{{v}^{2}}=mgh-mg2r=mg(h-2r)\] \[{{v}^{2}}=2g(h-2r)\] (i) Also, \[\frac{m{{v}^{2}}}{r}=xmg+mg\] Or \[{{v}^{2}}=(x+1)rg\] (ii) Equating Eqs. (i) and (ii), we get \[2g(h-2r)=(x+1)gr\] Or \[2gh=(x+1)gr+4gr=(x+5)gr\] \[h=\left( \frac{x+5}{2} \right)r\]
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