A bead - B - of mass m is attached to one end of a spring A B of natural length R and spring constant K-3-1 m g - R - The normal reaction at B just after it is released to move is-

The correct option is D mg Extension in the spring is x=AB R=2Rcos 30^∘ R =(√(3) 1)R Spring force F=kx=(√(3)+1)mg/R× (√(3) 1)R = 2mg From figure N = = (F + m ...

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Byju's Answer

Standard XII

Physics

Magnitude using Vector: Derivation

A bead ' B ' ...

Question

A bead 'B' of mass m is attached to one end of a spring(AB) of natural length R and spring constant K= ( $\sqrt{3}$ +1)mg/R. The normal reaction at B just after it is released to move is:

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Solution

The correct option is D
mg

Extension in the spring is

$x = A B - R = 2 R c o s 30^{\circ} - R$

$= (\sqrt{3} - 1) R$

Spring force

$F = k x = \frac{(\sqrt{3} + 1) m g}{R} \times (\sqrt{3} - 1) R$ = 2mg

From figure N = = (F + mg)cos $30^{\circ}$ = $\frac{\sqrt[x]{3} m g}{2}$

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A bead 'B' of mass m is attached to one end of a spring(AB) of natural length R and spring constant K= (√3+1)mg/R. The normal reaction at B just after it is released to move is:

The correct option is D mg Extension in the spring is x=AB−R=2Rcos 30∘−R =(√3−1)R Spring force F=kx=(√3+1)mgR×(√3−1)R = 2mg From figure N = = (F + mg)cos 30∘ = x√3 mg2

A bead 'B' of mass m is attached to one end of a spring(AB) of natural length R and spring constant K= ( $\sqrt{3}$ +1)mg/R. The normal reaction at B just after it is released to move is:

The correct option is D
mg

Extension in the spring is

$x = A B - R = 2 R c o s 30^{\circ} - R$

$= (\sqrt{3} - 1) R$

Spring force

$F = k x = \frac{(\sqrt{3} + 1) m g}{R} \times (\sqrt{3} - 1) R$ = 2mg

From figure N = = (F + mg)cos $30^{\circ}$ = $\frac{\sqrt[x]{3} m g}{2}$