Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Three identical particles A, B and C of mass $100 \mathrm{~kg}$ each are placed in a straight line with $\mathrm{A B=B C=13 \mathrm{~m}}$ . The gravitational force on a fourth particle $\mathrm{P}$ of the same mass is $\mathrm{F}$ , when placed at a distance $13 \mathrm{~m}$ from the particle $\mathrm{B}$ on the perpendicular bisector of the line $\mathrm{AC}$ . The value of F will be approximately :
Option: 1
$21\: \mathrm{G}$

Option: 2
$100\: \mathrm{G}$

Option: 3
$59\: \mathrm{G}$

Option: 4
$42\: \mathrm{G}$

Answers (1)

best_answer

Let, $\mathrm{F_{A},F_{B}\: and\:F_{c} }$ be the gravitational force on particle $\mathrm{P}$ by particle $\mathrm{A,B\, and\: C}$ respectively .

$\mathrm{\begin{aligned} \bar{F} &=\bar{F}_A+\bar{F}_B+\bar{F}_C \quad \text { (Given) } \\ \end{aligned}}$

$\mathrm{F =F_A \cos \theta+F_B+F_{C \cdot} \cos \theta }$

$\mathrm{=\frac{2 G(100)(100)}{r^2} \times \frac{13}{r}+\frac{G(100)^2}{13^2}}$

$\mathrm{r=\sqrt{\left ( 13 \right )^{2}+\left ( 13 \right )^{2}}}$

$\mathrm{F=\frac{2G\left ( 100 \right )^{2}\times 13}{\left ( 13 \right )^{2}\times 2\times 13\sqrt{2}}}+\mathrm{\frac{G\left ( 100 \right )^{2}}{13^{2}}}$

$=\mathrm{\frac{G\left ( 10 \right )^{4}}{\left ( 13 \right )^{2}}}\left [ \frac{1}{\sqrt{2}} +1\right ]$

$\mathrm{F=100\: G}$

Hence (2) is correct option

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

50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pK_b of ammonia solution is 4.75, the pH of the mixture will be :
Option: 1 3.75
Option: 2 4.75

Trending Articles/News

anam-khan

JEE Main 2025 response sheet errors: Delhi HC allows student to apply for JEE Adv, seeks NTA explanation

anam-khan

Four Kashmiri girls defy the odds, score 99+ percentile in JEE Main 2025

anam-khan

JEE, NEET, CUET coaching centres: Assam, Rajasthan have framed laws, but will they work?

Latest Question