Suppose a frictionless tunnel is made inside the earth along a diameter. A string of mass m and length R/2 is suspended in the tunnel with one end of the string at the surface of the earth. A force F is applied to pull the string slowly to the surface to the earth. Find the work done by the force F to pull the string completely on the surface of the earth. (Given acceleration due to gravity on the surface of the earth =g ,, radius of the earth =R )

Question

Suppose a frictionless tunnel is made inside the earth along a diameter. A string of mass m and length R/2 is suspended in the tunnel with one end of the string at the surface of the earth. A force F is applied to pull the string slowly to the surface to the earth. Find the work done by the force F to pull the string completely on the surface of the earth. (Given acceleration due to gravity on the surface of the earth =g ,, radius of the earth =R ) <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-lq5zakax3mqrmh8z.jpg" /> (A) (m g R/2) (B) (3 m g R/4) (C) (5 m g R/12) (D) (5 m g R/24)

Tardigrade Admin · Accepted Answer

Ui= displaystyle ∫ (.dm.)v= displaystyle ∫ R / 2R-(m/R / 2)d× (G M/2 R)(3 - (x2/R2)) =(29 G M m/24 R)=(29 g R2 m/24)=(29 m g R/24) Uf=-(G M m/R)=-(g R2 m/R)=-mgR W=Δ U=Uf-Ui=(5 m g R/24) <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-lq5zakax3mqrspqu.jpg" />

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

$\frac{3 m g R}{4}$

$\frac{5 m g R}{12}$

$\frac{5 m g R}{24}$

$U_{i} = \int (d m) v = \int_{R /2}^{R} - \frac{m}{R /2} d \times \frac{GM}{2 R} (3 - \frac{x ^{2}}{R ^{2}})$
$= \frac{29 GM m}{24 R} = \frac{29 g R ^{2} m}{24} = \frac{29 m g R}{24}$
$U_{f} = - \frac{GM m}{R} = - \frac{g R ^{2} m}{R} = - m g R$
$W = Δ U = U_{f} - U_{i} = \frac{5 m g R}{24}$

2mgR​

43mgR​

125mgR​

245mgR​

Ui​=∫(dm)v=∫R/2R​−R/2m​d×2RGM​(3−R2x2​) =24R29GMm​=2429gR2m​=2429mgR​ Uf​=−RGMm​=−RgR2m​=−mgR W=ΔU=Uf​−Ui​=245mgR​

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

$\frac{3 m g R}{4}$

$\frac{5 m g R}{12}$

$\frac{5 m g R}{24}$

$U_{i} = \int (d m) v = \int_{R /2}^{R} - \frac{m}{R /2} d \times \frac{GM}{2 R} (3 - \frac{x ^{2}}{R ^{2}})$
$= \frac{29 GM m}{24 R} = \frac{29 g R ^{2} m}{24} = \frac{29 m g R}{24}$
$U_{f} = - \frac{GM m}{R} = - \frac{g R ^{2} m}{R} = - m g R$
$W = Δ U = U_{f} - U_{i} = \frac{5 m g R}{24}$