All surfaces shown in the figure are smooth. The system is released with the spring unstretched. In equilibrium, compression in the spring will be -

Question

All surfaces shown in the figure are smooth. The system is released with the spring unstretched. In equilibrium, compression in the spring will be - <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-yvkp0jvygax3vl2w.jpg" /> (A) (m g/√2 k) (B) (2 m g/k) (C) ((M + m) g/√2 k) (D) (m g/k)

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Let N be the normal reaction between m and M , Equilibrium of M N sin 45° =kx ... (i) Equilibrium of m in vertical direction gives N cos 45° =mg ... (ii) From Eqs. (i) and (ii), we get x=(m g/k)

Q. All surfaces shown in the figure are smooth. The system is released with the spring unstretched. In equilibrium, compression in the spring will be -

$\frac{m g}{2 k}$

$\frac{2 m g}{k}$

$\frac{( M + m ) g}{2 k}$

$\frac{m g}{k}$

Let $N$ be the normal reaction between $m$ and $M$ ,
Equilibrium of $M$
$N s in 4 5^{^{\circ}} = k x$ ... (i)
Equilibrium of $m$ in vertical direction gives
$N cos 4 5^{^{\circ}} = m g$ ... (ii)
From Eqs. (i) and (ii), we get
$x = \frac{m g}{k}$

Q. All surfaces shown in the figure are smooth. The system is released with the spring unstretched. In equilibrium, compression in the spring will be -

2​kmg​

k2mg​

2​k(M+m)g​

kmg​

Let N be the normal reaction between m and M , Equilibrium of M Nsin45∘=kx ... (i) Equilibrium of m in vertical direction gives Ncos45∘=mg ... (ii) From Eqs. (i) and (ii), we get x=kmg​

$\frac{m g}{2 k}$

$\frac{2 m g}{k}$

$\frac{( M + m ) g}{2 k}$

$\frac{m g}{k}$

Let $N$ be the normal reaction between $m$ and $M$ ,
Equilibrium of $M$
$N s in 4 5^{^{\circ}} = k x$ ... (i)
Equilibrium of $m$ in vertical direction gives
$N cos 4 5^{^{\circ}} = m g$ ... (ii)
From Eqs. (i) and (ii), we get
$x = \frac{m g}{k}$