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Mass of the Earth, M = 6.0 $*{10}^{24}$ kg

Radius of the Earth, R = 6400 km = 6.4 $*{10}^6$ m

Height of geostationary satellite from the surface of the Earth, h = 36000 km = 3.6 $*{10}^7$ m

Gravitational potential energy due to Earth's gravity at height h:

= $\frac{-GM}{(R+h)}$

= $-\frac{6.67*{10}^{-11}*6.0*{10}^{24}}{3.6*{10}^7+6.4*{10}^6}$ = $-\frac{6.67*6.0*{10}^{13}}{4.24*{10}^7}$ = $-$ 9.44 $*{10}^6$ J/kg

Mass of each sphere, M = 100 kg

Separation between the spheres, r = 1 m

X is the midpoint between the spheres.

Gravitational force at point x will be zero. This is because gravitational force exerted by each spheres will act in opposite directions.

Gravitational potential at point x:

= $\frac{-GM}{\frac{r}{2}}$ $\frac{-GM}{\frac{r}{2}}$ = $-$ 4 $\frac{GM}{r}$ = $-\frac{4*6.67*{10}^{-11}*100}{1}$ = $-2.668*{10}^{-8}$ J/kg

Any object placed at point x will be in equilibrium state, but the equilibrium is unstable. This is because any change in the position of the object will change the effective force in that direction.

ASBM University has not yet released the placement statistics specifically applying to BTech. However, interested candidates can refer to the student reviews posted on the web to find out the highest package offered to BTech students at ASBM University. As per the latest review posted by a BTech student on Shiksha.com, the highest salary package at ASBMU is INR 10 LPA. The student gave a five-star rating (out of five) to BTech placements at ASBMU.

Escape velocity of the projectile on the Earth's surface, $v_{esc}$ = 11.2 km/s = 11.2 ${*10}^3$ m/s

Projection velocity of the projectile = 3 $v_{esc}$

Mass of the projectile = m

Velocity of the projectile far away from the Earth = $v_f$

Total energy of the projectile on the Earth = $\frac{1}{2}$ m ${v_p}^2$ $-$ $\frac{1}{2}$ m ${v_{esc}}^2$

Gravitational potential energy of the projectile far away from the Earth is zero.

Total energy of the projectile far away from the Earth = $\frac{1}{2}$ m ${v_f}^2$

From the law of conservation of energy

$\frac{1}{2}$ m ${v_p}^2$ $–$ $\frac{1}{2}$ m ${v_{esc}}^2$ = $\frac{1}{2}$ m ${v_f}^2$

${v_p}^2-{v_{esc}}^2={v_f}^2$

$v_f$ = $\sqrt{{v_p}^2-{v_{esc}}^2}$ = $\sqrt{{{(3v}_{esc})}^2-{v_{esc}}^2}$ = $\sqrt{8*}$ 11.2= 31.68 km/s

For BBA at Thapar (School of Management), the total fee is roughly ₹8–9 Lacs for the full programme (can vary a bit by year).
The average package so far is around ₹6–7 LPA, with top roles going higher as the school's placements mature.

It's a growing program, but Thapar's brand + campus exposure definitely helps.

While most universities in Nepal do not clear submission of IELTS score, Indian students are advised to take up exam to meet English language requirements. Listed below are the minimum score requirements:

Weight of a body of mass m at Earth's surface, W = mg = 250 N

Body of mass m is located at depth, d = $\frac{1}{2}{R}_e$ , where $R_e$ is radius of the Earth

Acceleration due to gravity at depth g (d) is given by : g' = (1 - $\frac{d}{R_e}$ )g = (1/2)g

Weight of the body at depth d

W' = mg' = (1/2) mg = (1/2)W = (1/2) x 250 N = 125 N

The tuition fee for the BBA programme at NPU Jharkhand is approximately INR 26,250. This fee may include tuition and other components like security deposit or miscellaneous charges. Additional costs for hostel, mess, and materials may apply. Students should confirm fee details from the official university source during admission. 

Weight of the body, W = 63 N

Acceleration due to gravity at h from the Earth's surface is given by

g' = $\frac{g}{{\left(\frac{1+h}{R_e}\right)}^2}$ where g = acceleration due to gravity on the Earth's surface,  $R_e$ = Radius of the Earth. h = $\frac{R_e}{2}$

g' = $\frac{g}{{\left(\frac{1+R_e/2}{R_e}\right)}^2}$ = $\frac{g}{{\left(\frac{1+h}{2h}\right)}^2}$ = (4/9)g

Weight of the body of mass m at a height h is given by

W' = m X g' = (4/9) mg = (4/9) x w = (4/9) x 63 N = 28 N

JIET Group of Institutions does not take its own entrance exam but accepts national-level exams such as JEE Main, CAT, and GATE. Difficulty varies with the exam you sit for. If you prepare well as per the exam's syllabus, your chances are strong. Mock tests and last year's question papers are useful.

Indian students must meet admission requirements mentioned below for MBBS in Nepal-

• A medical fitness certificate from a registered clinic or hospital.
• A minimum of 50 - 60% aggregate marks in the above-mentioned subjects;
• Senior Secondary Higher Education with Physics, Chemistry, and Biology;
• NEET-UG entrance exam qualification;
• Must be at least 17 years of age before December end of the admission year; and
• Valid passport with at least 18 months of validity

Note: Some universities may require the Medical Education Commission Entrance Examination (MECEE) in addition to NEET.

Yes, admission to the BBA course at NPU Jharkhand is generally merit-based, determined by the candidate's performance in their Class 12 board exams. As of now, no entrance exam is required, making it possible to secure direct admission if minimum eligibility conditions are met.

Distance of the Earth from the Sun, $r_e$ = 1.50 × 10⁸ km = 1.5 × 10¹¹ m

Time period of the Earth = $T_e$ , Time period of the Saturn $T_s$ = ${29.5T}_e$

Distance of Saturn from the Sun = $r_s$

From Kepler’s 3rd law of planetary motion, we have T =( ${\frac{4{\pi }^2{r}^3}{GM})}^{1/2}$

For Saturn and Sun, we can write, $\frac{r_s^3}{r_e^3}$ = $\frac{T_s^2}{T_e^3}$

$r_s$ = $r_e({\frac{29.5T_e}{T_e})}^{2/3}$ = 1.5 × 10¹¹$\times ({\frac{29.5}{1})}^{2/3}$ = 1.4 X ${10}^{12}$ m