(B) When the block is lowered slowly,the equilibrium position is reached when the spring force equals the gravitational force: $k d = m g$,which impli

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(B) When the block is lowered slowly,the equilibrium position is reached when the spring force equals the gravitational force: $k d = m g$,which implies $d = \frac{m g}{k}$.When the block is allowed to fall suddenly from the unstretched position,the block undergoes simple harmonic motion. At the maximum extension $x$,the change in gravitational potential energy equals the elastic potential energy stored in the spring:$m g x = \frac{1}{2} k x^2$Solving for $x$ (where $x \neq 0$):$x = \frac{2 m g}{k}$Since $d = \frac{m g}{k}$,we substitute this into the equation:$x = 2 d$.

Class 11 Chemistry Data Entry - Chemistry Level 0 of all question bank

To the free end of a spring hanging from a rigid support,a block of mass $m$ is hung and slowly allowed to come to its equilibrium position. Then the stretching in the spring is $d$. If the same block is attached to the same spring and allowed to fall suddenly,the amount of stretching is: (force constant,$k$)

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A
$\frac{m g}{k}$
B
$2 d$
C
$\frac{m g}{3 k}$
D
$4 d$

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Consider the set of all triangles $OPQ$ where '$O$' is the origin and $P$,$Q$ are distinct points in the plane with non-negative integral coordinates $(x, y)$ such that $5x + y = 99$. The number of such distinct triangles whose area is a positive integer is:

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