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Chapter 6: Problem 681

Two point masses \(\mathrm{A}\) and \(\mathrm{B}\) having masses in the ratio $4: 3\( are separated by a distance of \)\operatorname{lm}$. When another point mass of mass \(\mathrm{M}\) is placed in between \(\mathrm{A}\) and \(\mathrm{B}\) the forces \(\mathrm{A}\) and is \((1 / 3 \mathrm{rd})\) of the force between \(\mathrm{B}\) and \(\mathrm{C}\), Then the distance \(\mathrm{C}\) from \(\mathrm{A}\) is \(=\ldots \ldots \ldots \mathrm{m}\) (A) \((2 / 3)\) (B) \(1 / 3\) (C) \(1 / 4\) (D) \(2 / 7\)

Short Answer

Expert verified
The distance between A and C is 2/7 meters. Therefore, the correct answer is (D) 2/7.

Step by step solution

01

Let mA be the mass of A and mB the mass of B. Since their masses are in the ratio 4:3, we can write mA = 4m and mB = 3m, where m is some arbitrary mass unit. #Step 2: Write down the forces between A and C, and B and C#

According to the universal law of gravitation, the force F between two point masses is given by \[F = G\frac{m_{1}m_{2}}{r^2}\], where G is the gravitational constant, m1 and m2 are the masses and r is the distance between them. Therefore, the forces between A and C, and B and C can be written as \[F_{AC} = G\frac{mA \times M}{x^2}\] and \[F_{BC} = G\frac{mB \times M}{(1-x)^2}\], where M is the mass of point C and x is the distance between A and C. #Step 3: Formulate the given condition#
02

We are given that the force between A and C is 1/3 of the force between B and C, so we can write the condition as \[F_{AC} = \frac{1}{3}F_{BC}\]. Substitute the expressions for F_AC and F_BC from Step 2 and we get \[\frac{m_{A}\times M}{x^{2}} = \frac{1}{3}\frac{m_{B} \times M}{(1-x)^{2}}\] #Step 4: Solve for the distance x#

We can cancel out the mass M in the equation, and then substitute mA = 4m and mB = 3m from step 1, which gives us: \[\frac{4m}{x^2} = \frac{1}{3}\frac{3m}{(1-x)^2}\]. After canceling out the mass m and solving for x, we get: \[4(1-x)^2 = 3x^2 \Rightarrow 4 - 8x + 4x^2= 3x^2 \Rightarrow x^2 = 8x - 4 \Rightarrow x = \frac{2}{7}\] So, the distance between A and C is x = 2/7 meters. Therefore, the correct answer is (D) 2/7.

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Most popular questions from this chapter

A geo-stationary satellite is orbiting the earth of a height of \(6 \mathrm{R}\) above the surface of earth, \(\mathrm{R}\) being the radius of earth. The time period of another satellite at a height of \(2.5 \mathrm{R}\) from the surface for earth is \(=\ldots \ldots \ldots \ldots\) (A) 6 (B) \(6 \sqrt{2}\) (C) 10 (D) \(6 / \sqrt{2}\)

At what distance from the center of earth, the value of acceleration due to gravity \(g\) will be half that of the surfaces \((\mathrm{R}=\) Radius of earth \()\) (A) \(2 \mathrm{R}\) (B) \(\mathrm{R}\) (C) \(1.414 \mathrm{R}\) (D) \(0.414 \mathrm{R}\)

A particle of mass \(\mathrm{M}\) is situated at the center of a spherical shell of same mass and radius a the magnitude of gravitational potential at a point situated at (a/2) distance from the center will be (A) \([(4 \mathrm{GM}) / \mathrm{a}]\) (B) \((\mathrm{GM} / \mathrm{a})\) (C) \([(2 \mathrm{GM}) / \mathrm{a}]\) (D) \([(3 \mathrm{GM}) / \mathrm{a}]\)

let \(\mathrm{V}\) and \(\mathrm{E}\) denote the gravitational potential and gravitational field at a point. Then the match the following \(\begin{array}{ll}\text { Table }-1 & \text { Table }-2\end{array}\) (A) \(\mathrm{E}=0, \mathrm{~V}=0\) (P) At center of spherical shell (B) \(\mathrm{E} \neq 0, \mathrm{~V}=0\) (Q) At center of solid sphere (C) \(\mathrm{V} \neq 0, \mathrm{E}=0\) (R) at centre of circular ring (D) \(\mathrm{V} \neq 0, \mathrm{E} \neq 0\) (S) At centre of two point masses of equal magnitude (T) None

Two bodies of masses \(m_{1}\) and \(m_{2}\) are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual a gravitational attraction Their relative velocity of approach at separation distance \(\mathrm{r}\) between them is (A) $\left[\left\\{2 \mathrm{G}\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right)\right\\} / \mathrm{r}\right]^{-1 / 2}$ (B) $\left[\left\\{2 \mathrm{G}\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)\right\\} / \mathrm{r}\right]^{1 / 2}$ (C) $\left[\mathrm{r} /\left\\{2 \mathrm{G}\left(\mathrm{m}_{1} \mathrm{~m}_{2}\right)\right\\} / \mathrm{r}\right]^{1 / 2}$ (D) $\left[\left(2 \mathrm{Gm}_{1} \mathrm{~m}_{2}\right) / \mathrm{r}\right]^{1 / 2}$
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