Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

A straight rod of length L extends from x = a to x = L + a. The gravitational force it exerts on a point mass 'm' at x = 0, if the mass per unit length of the rod is A + Bx² , is given by :
Option 1)
$Gm \left [ A \left ( \frac{1}{a + L} - \frac{1}{a} \right ) - BL \right ]$
Option 2)
$Gm \left [ A \left ( \frac{1}{a } - \frac{1}{a +L} \right ) - BL \right ]$
Option 3)
$Gm \left [ A \left ( \frac{1}{a} - \frac{1}{a + L} \right ) + BL \right ]$
Option 4)
$Gm \left [ A \left ( \frac{1}{a + L} - \frac{1}{a} \right ) + BL \right ]$

Answers (1)

best_answer

Newton's Law of Gravitation -

$F\; \alpha\; \frac{m_{1}m_{2}}{r^{2}}$

$F\; = \frac{G\, m_{1}\, m_{2}}{r^{2}}$

$F\rightarrow$ Force

$G\rightarrow$ Gravitalional constant

$m_1,m_2\rightarrow$ Masses

$r\rightarrow$ Distance between masses

- wherein

Force is along the line joining the two masses

$\lambda =A+Bx^{2}$

assume dx element at a distance x from $x=0$

So $dm=\lambda dx$

$dm=\left ( A+Bx^{2} \right )dx$

$dF=\frac{Gm_{1}dm}{x^{2}}$

$F=\int_{a}^{a+L}\frac{Gm_{1}}{x^{2}}\left ( A+Bx^{2} \right )dx$

$F=Gm\left ( \int_{a}^{a+L}\frac{A}{x^{2}}dx+\int_{a}^{a+L}B\, \, dx \right )$

$=Gm\left [ \frac{-A}{x}+Bx \right ]_{a}^{L+a}$ $F=Gm\left [ A\frac{1}{a}-\frac{1}{a+L}+BL\right ]$

Option 1)

$Gm \left [ A \left ( \frac{1}{a + L} - \frac{1}{a} \right ) - BL \right ]$

Option 2)

$Gm \left [ A \left ( \frac{1}{a } - \frac{1}{a +L} \right ) - BL \right ]$

Option 3)
$Gm \left [ A \left ( \frac{1}{a} - \frac{1}{a + L} \right ) + BL \right ]$
Option 4)

$Gm \left [ A \left ( \frac{1}{a + L} - \frac{1}{a} \right ) + BL \right ]$

Posted by

admin

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today

anam-khan

JoSAA round 2 cut-off 2025 out; BTech closing ranks for top engineering courses at IITs

anam-khan

Goa Engineering Admission 2025: 1,415 eligible for BE seats; merit list on June 27

Latest Question