'M' $\mathrm{kg}$ weight is suspended from a weightless spring and it has period ' $T^{\prime}$. If now ' $4 \mathrm{~m}^{\prime} \mathrm{kg}$ weight is suspended from the same spring, the new period will beA. $\frac{T}{2}$B. $2 \mathrm{~T}$C. $4 \mathrm{~T}$D. $\mathrm{T}$

‘M’ $\mathrm{kg}$ weight is suspended from a weightless spring and it has period ‘ $T^{\prime}$. If now ‘ $4 \mathrm{~m}^{\prime} \mathrm{kg}$ weight is suspended from the same spring, the new period will beA. $\frac{T}{2}$B. $2 \mathrm{~T}$C. $4 \mathrm{~T}$D. $\mathrm{T}$

‘M’ $\mathrm{kg}$ weight is suspended from a weightless spring and it has period ‘ $T^{\prime}$. If now ‘ $4 \mathrm{~m}^{\prime} \mathrm{kg}$ weight is suspended from the same spring, the new period will be
A. $\frac{T}{2}$
B. $2 \mathrm{~T}$
C. $4 \mathrm{~T}$
D. $\mathrm{T}$

Physics

‘M’ $\mathrm{kg}$ weight is suspended from a weightless spring and it has period ‘ $T^{\prime}$. If now ‘ $4 \mathrm{~m}^{\prime} \mathrm{kg}$ weight is suspended from the same spring, the new period will be
A. $\frac{T}{2}$
B. $2 \mathrm{~T}$
C. $4 \mathrm{~T}$
D. $\mathrm{T}$

Correct answer is B.

$\begin{array}{l}
\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}} \\
\mathrm{T}_{1}=\mathrm{T} \\
\mathrm{m}_{1}=1 \mathrm{~kg} \\
\mathrm{~m}_{2}=4 \mathrm{~kg} \\
\mathrm{~T}=2 \pi \sqrt{\frac{1}{\mathrm{k}}} \\
\mathrm{T}_{2}=2 \pi \sqrt{\frac{4}{\mathrm{k}}} \\
\mathrm{T}_{2}=2 \mathrm{~T}
\end{array}$

i

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