The tension in a string holding a solid block below the surface of a liquid (where ρ text liquid ρ text block ) as in shown in the figure is T0 when the system is at rest. Then what will be the tension in the string if the system has upward acceleration a ?

Question

The tension in a string holding a solid block below the surface of a liquid (where ρ text liquid > ρ text block ) as in shown in the figure is T0 when the system is at rest. Then what will be the tension in the string if the system has upward acceleration a ? (A) T0(1-(a/g)) (B) T0(1+(a/g)) (C) T0((a/g)-1) (D) (a/g) T0

Tardigrade Admin · Accepted Answer

Let v be the volume of the solid block of density ρ. Let ρ1 be the density of water. Weight of body =v ρ g. When the body is immersed in water, Tension in the string = Upward thrust - Weight of the body ⇒ T=v ρ1 g-v ρ g=v g(ρ1-ρ2) When the lift is moving upwards with acceleration a, the tension in the string is T=v(ρ1-ρ)(g+a) From Eqs. (i) and (ii), T=T0(1+a / g).

Q. The tension in a string holding a solid block below the surface of a liquid (where $ρ_{liquid} > ρ_{block}$ ) as in shown in the figure is $T_{0}$ when the system is at rest. Then what will be the tension in the string if the system has upward acceleration $a$ ?

$T_{0} (1 - \frac{a}{g})$

$T_{0} (1 + \frac{a}{g})$

$T_{0} (\frac{a}{g} - 1)$

$\frac{a}{g} T_{0}$

Q. The tension in a string holding a solid block below the surface of a liquid (where ρliquid ​>ρblock ​ ) as in shown in the figure is T0​ when the system is at rest. Then what will be the tension in the string if the system has upward acceleration a ?

T0​(1−ga​)

T0​(1+ga​)

T0​(ga​−1)

ga​T0​

Q. The tension in a string holding a solid block below the surface of a liquid (where $ρ_{liquid} > ρ_{block}$ ) as in shown in the figure is $T_{0}$ when the system is at rest. Then what will be the tension in the string if the system has upward acceleration $a$ ?

$T_{0} (1 - \frac{a}{g})$

$T_{0} (1 + \frac{a}{g})$

$T_{0} (\frac{a}{g} - 1)$

$\frac{a}{g} T_{0}$