$A$ massless spring gets elongated by an amount $x_1$ under a tension of $5 \ N$. Its elongation is $x_2$ under a tension of $7 \ N$. For an elongation of $(5x_1 - 2x_2)$,the tension in the spring will be: (in $N$)

Two blocks each of mass $m$ are connected to a spring of spring constant $k.$ If both are given velocity $v$ in opposite directions,then the maximum elongation of the spring is

$A$ spring of force constant $k$ is cut into two equal halves. The force constant of each half is

Two massless springs of spring constant $K_1$ and $K_2$ are connected in series,suspended vertically,and a certain mass is attached to the free end. If $e_1$ and $e_2$ are their respective extensions and $f$ is the stretching force,the total extension produced is:

$A$ mass of $10 \ kg$ is suspended from a spring balance. It is pulled aside by a horizontal string so that it makes an angle of $60^{\circ}$ with the vertical. The new reading of the balance is

$A$ body of mass $m$ is tied to one end of a spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is $1 \, cm$. If the angular velocity is doubled,the elongation in the spring is $5 \, cm$. The original length of the spring is ............ $cm$.