Three masses 200 kg, 300 kg and 400 kg are placed at the vertices of an equilateral triangle with sides 20 m. They are rearranged on the vertices of a bigger triangle of side 25 m and with the same centre. The work done in this process ____ J. (Gravitational constant G = 6.7 times 10^-11 text N m^2 / textkg^2$G = 6.7 \times 10^{-11} \text{ N m}^{2} / \text{kg}^{2}$)

### Related Formula E = frac-GM_E m2r$$E = \frac{-GM_E m}{2r}$$ Delta E = E_f - E_i$$\Delta E = E_f - E_i$$ g = fracGM_ER_E^2 implies GM_E = g R_E^2$$g = \frac{GM_E}{R_E^2} \implies GM_E = g R_E^2$$ ### Core Logic Energy of a satellite in a circular orbit is given as E = frac-GM_E m2r$E = \frac{-GM_E m}{2r}$ where r is the radius of the circular orbit. Required energy to be supplied Delta E = E_f - E_i$\Delta E = E_f - E_i$: Delta E = left( frac-GM_E m2(3R_E) right) - left( frac-GM_E m2(1.5R_E) right)$$\Delta E = \left( \frac{-GM_E m}{2(3R_E)} \right) - \left( \frac{-GM_E m}{2(1.5R_E)} \right)$$ Delta E = frac-GM_E m6R_E + fracGM_E m3R_E = fracGM_E m6R_E$$\Delta E = \frac{-GM_E m}{6R_E} + \frac{GM_E m}{3R_E} = \frac{GM_E m}{6R_E}$$ ### Step 1: Evaluate Delta E Substitute GM_E = g R_E^2$GM_E = g R_E^2$ into the expression: Delta E = frac(g R_E^2) m6R_E = frac16 m g R_E$$\Delta E = \frac{(g R_E^2) m}{6R_E} = \frac{1}{6} m g R_E$$ = frac16 times 100 times 10 times (6 times 10^6)$$= \frac{1}{6} \times 100 \times 10 \times (6 \times 10^6)$$ = 1000 times 10^6text J$$= 1000 \times 10^6\text{ J}$$ Comparing with alpha times 10^6text J$\alpha \times 10^6\text{ J}$, we get alpha = 1000$\alpha = 1000$. ### Pattern Recognition Orbital transition energy is always Delta E = fracGMm2 (frac1r_i - frac1r_f)$\Delta E = \frac{GMm}{2} (\frac{1}{r_i} - \frac{1}{r_f})$. Never forget the factor of 2 in the denominator (which accounts for kinetic energy contribution in orbit). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation

### Related Formula V_e = sqrtfrac2GMR = sqrtfrac2G rho frac4pi R^33R implies V_e propto sqrtrho times R$$V_e = \sqrt{\frac{2GM}{R}} = \sqrt{\frac{2G \rho \frac{4\pi R^3}{3}}{R}} \implies V_e \propto \sqrt{\rho} \times R$$ ### Core Logic Given density and radius of B are 10% (0.1$0.1$) of planet A: frac(V_e)_B(V_e)_A = sqrtfracrho_Brho_A times fracR_BR_A = sqrt0.1 times 0.1 = frac110sqrt10$$\frac{(V_e)_B}{(V_e)_A} = \sqrt{\frac{\rho_B}{\rho_A}} \times \frac{R_B}{R_A} = \sqrt{0.1} \times 0.1 = \frac{1}{10\sqrt{10}}$$ (V_e)_B = frac10 times 100010sqrt10 = 100sqrt10 text m/sec$$(V_e)_B = \frac{10 \times 1000}{10\sqrt{10}} = 100\sqrt{10} \text{ m/sec}$$ ### Pattern Recognition Sees: Escape velocity scaling with density and radius. Shortcut: Express escape velocity in terms of density rho$\rho$ and radius R$R$, then take ratios. Check: Matches option (3). ✓ ### Chapter Mix Class 11 Physics: Gravitation

### Related Formula F = fracG m_1 m_2r^2$$F = \frac{G m_1 m_2}{r^2}$$ ### Core Logic

Initial configuration: F_1 = 2sqrt2 fracGmm_0r^2$| New configuration (after interchanging 3M and 4M): $$F_1 = 2\sqrt{2} \frac{Gmm_0}{r^2}$| New configuration (after interchanging 3M and 4M): $$F_2 = \sqrt{10} \frac{Gmm_0}{r^2}Ratio:$Ratio:$\frac{F_1}{F_2} = \frac{2\sqrt{2}}{\sqrt{10}} = \frac{2}{\sqrt{5}} \implies \alpha = 2$$ ### Pattern Recognition Sees: Gravitational force vector sum at center of square. Shortcut: Compute net vector resultants for both mass configurations and take ratio. Check: Matches option (1). ✓ ### Chapter Mix Class 11 Physics: Gravitation

### Related Formula T = 2pi sqrtfracR^3GM$$T = 2\pi \sqrt{\frac{R^3}{GM}}$$ M = rho cdot frac43pi R^3$$M = \rho \cdot \frac{4}{3}\pi R^3$$ g = fracGMR^2$$g = \frac{GM}{R^2}$$ ### Core Logic Evaluating Statement I: Substituting total mass M = rho cdot frac43pi R^3$M = \rho \cdot \frac{4}{3}\pi R^3$ into time period formula: T = 2pi sqrtfracR^3G left(rho cdot frac43pi R^3right) = sqrtfrac3piG rho$$T = 2\pi \sqrt{\frac{R^3}{G \left(\rho \cdot \frac{4}{3}\pi R^3\right)}} = \sqrt{\frac{3\pi}{G \rho}}$$ Hence, time period depends inversely on the square root of Earth's density rho$\rho$. Statement I is true. Evaluating Statement II: Substituting g = fracGMR_e^2$g = \frac{GM}{R_e^2}$ into T = 2pi sqrtfracR_e^3GM$T = 2\pi \sqrt{\frac{R_e^3}{GM}}$: T = 2pi sqrtfracR_eg$$T = 2\pi \sqrt{\frac{R_e}{g}}$$ Statement II is also true. ### Step 1: Final Conclusion Both Statement I and Statement II are true. ### Pattern Recognition Near-surface satellite shortcut: T = sqrt3pi / (Grho) approx 84.6 text min$T = \sqrt{3\pi / (G\rho)} \approx 84.6 \text{ min}$. Shows direct dependence on Earth's density rho$\rho$ and formula T = 2pi sqrtR_e / g$T = 2\pi \sqrt{R_e / g}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation