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BITSAT 2014
Exam
BITSAT
Year
2014
Subject
Physics
Chapter
Electrostatics

BITSAT 2014 · Physics · Electrostatics

A spherically symmetric gravitational system of particles has a...

A spherically symmetric gravitational system of particles has a mass density \rho=\left\{ρ0 for r≤R0 for r>R\begin{array}{ll}\rho_{0} & \text{ for } \mathrm{r} \leq \mathrm{R} \\ 0 & \text{ for } \mathrm{r} > \mathrm{R}\end{array}\right. where r0\mathrm{r}_{0} is a constant. A test mass can undergo circular motion under the influence of the gravitational field of particles. Its speed V\mathrm{V} as a function of distance r(0<r<∞)r(0< r< \infty) from the centre of the system is represented byA spherically symmetric gravitational system of particles has a mass density =\ array ll _ 0 & for r ≤ R 0 & for r > RA spherically symmetric gravitational system of particles has a mass density =\ array ll _ 0 & for r ≤ R 0 & for r > RA spherically symmetric gravitational system of particles has a mass density =\ array ll _ 0 & for r ≤ R 0 & for r > RA spherically symmetric gravitational system of particles has a mass density =\ array ll _ 0 & for r ≤ R 0 & for r > R

Options

  1. AAmong the element Li, N, C and Be , one with the largest atomic radius is
  2. BAmong the element Li, N, C and Be , one with the largest atomic radius is
  3. CAmong the element Li, N, C and Be , one with the largest atomic radius is
  4. DAmong the element Li, N, C and Be , one with the largest atomic radius is

Correct answer

C. Among the element Li, N, C and Be , one with the largest atomic radius is

Step-by-step solution

Given mass density P={P0) for (r≤RO)for(r>R})Sincegravitationaldependsonmass,whichiszeroforP=\left\{P_{0}\right.)\text{ for } (r \leq R O) for (\left.r > R\right\}) Since gravitational depends on mass, which is zero forr > R,henceitonlyactsfor(r≤R)., hence it only acts for (r \leq R).\begin{aligned}&\overrightarrow{\mathrm{F}_{\text{graw }}=\overrightarrow{\mathrm{F}_{\text{circular }}\\ &\Rightarrow \frac{\mathrm{GM}_{\text{sphere }} \times \mathrm{M}}{\mathrm{R}^{2}}=\frac{\mathrm{M}(\mathrm{e})^{2}}{\mathrm{R}}\\ &\Rightarrow \mathrm{V}^{2} \alpha \frac{\mathrm{M}_{\text{sphere }}}{\mathrm{R}} \text{ since } \mathrm{r} \leq \mathrm{R}\\ &\Rightarrow \mathrm{V}^{2} \alpha \frac{\mathrm{R}^{3}}{\mathrm{R}} \alpha \mathrm{R}^{2} \Rightarrow \mathrm{V} \alpha \mathrm{R} \quad \text{ straight line }+\mathrm{ve} \text{ slope } \end{aligned}$Beyond(Beyond (\mathrm{r}=\mathrm{R}$), mass density (=0 \Rightarrow) acceleration due to gravity (=0 \Rightarrow) velocity stays constant.
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