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Power= Voltage x current= 2.5*750*10^-3= 1.875 watt

power=VI=3*600*10^-3=1800*10^-3=1.8Watt

P=I^2RP=I×I×RP= VI V=3V I=6×10^3AP=3×6×10^3P=1.8×10^4W

R=V/I V=3V I=600×10^3A 3/(600×10^3) 3/6×10^5 0.5×10^-5ohms

E=I^2×R×T I×I×R×T I×V×T I=6×10^5A V=3V T= 4h =(4×60×60)seconds =1.44×10^4seconds 6×10^5×3×1.44×10^4 25.92×10^9×1.44×10^4 2.592×10^4J

Given :V=2.5 VI=750mA= 750x10^-3 Apower =?

Power = V*I =2.5x750x10^-3=1.875W

Resistance :V=IRR=V/I=2.5/750x10^- 3 = 3.3 ohms

Energy consumed if bulb is lighted for 4 hours :E=P*T=1.875x4=7.500Wh=27x 10^3 joules

Option A

(1) P1=100 WV1 = 220 vP = V×IP = V × (V/R)P = V^2/R100 = (220×220)/R1R1 = 484 ohm(2) P2 = 60 WV2= 220 V60 = (220×220)/R2R2 = 806.6 ohmHence R2 is greater than R1

Resistances of both the bulbs are R1=V^2/P1 = 220^2/25 R2 = V^2/P2 = 220^2/100 Hence R1gt ; R2R1gt ; R2 When connected in series, the voltages divide in them in the ratio of their resistances. The voltage of 440 V devides in such a way that voltage across 25 w bulb will be more than 220 V. Hence 25 w bulb will fuse.

Assume- h= 3200km = 3200000m = 3.2 x10^6 R= 6400km = 6400000m = 6.4 x 10^6 g= 9.8m/s^2We know that- g'= g[R^2/(R+h)^2] = 9.8(40.96 x 10^12/92.16 x 10^12) = 9.8 x 4/9 = 4.355m/s^2 ~ 4.4m/s^2

Thus, value of g at a place 3200km above surface of the earth is 4.4m/s^2.

v good

h= 3200km = 3200000m = 3.2 x10^6 R= 6400km = 6400000m = 6.4 x 10^6 g= 9.8m/s^2We know that- g'= g[R^2/(R+h)^2] = 9.8(40.96 x 10^12/92.16 x 10^12) = 9.8 x 4/9 = 4.355m/s^2 ~ 4.4m/s^2

Thus, value of g at a place 3200km above surface of the earth is 4.4m/s^2.

thank you

Assume- h= 3200km = 3200000m = 3.2 x10^6 R= 6400km = 6400000m = 6.4 x 10^6 g= 9.8m/s^2We know that- g'= g[R^2/(R+h)^2] = 9.8(40.96 x 10^12/92.16 x 10^12) = 9.8 x 4/9 = 4.355m/s^2 ~ 4.4m/s^2

Thus, value of g at a place 3200km above surface of the earth is 4.4m/s^2.

Given here:

F∝1/Ror,F =kRNow, when one of the particles goes around the other then it is acted upon by a centripetal force given by:

Fcentri= mv2/Rnow,

mv2/R = kR

v =R√(k/m)orv∝ R

The gravitational force on M due to the three masses will form a system of concurrent forces of equal magnitude GmM/(x^2). Since these forces are at an angle of 120 degrees with respect to each other, the net resultant of force will be zero.

Hence, Option A is correct.

G' = g(1 -2h/R) rearrange this expression you getg'=g-2hg/Rg'-g=-2hg/Rby dividing g on both sideswe get

(g'-g)/g =-2h/R

-2/100= -2h/R

Or

h = R/100 = 6400/100 = 64 km

Initial speed of the body is, u = 10 km/s = 10000 m/s

Total energy of the body on the surface of earth is, Ei= PE + KE

=> Ei= -GMm/R+ ½ mu2

[m is the mass of the body]

Suppose it reaches a height ‘h’ from the surface of the earth. Its total energy at that height is,

Ef= -GMm/(R + h)

[at this height the body has no KE]

By law of conservation of energy,

Ei= Ef

=> -GMm/R + ½ mu2= -GMm/(R + h)

=> ½ u2= GM/R – GM/(R + h)

=> ½ u2= GM[R+h-R]/{R(R+h)}

=> ½ u2= GMh/(R2+Rh)

=> (R2+Rh)u2= 2GMh

=> R2u2+ Ru2h = 2GMh

=> h = R2u2/(2GM – Ru2)

Here, R = 6400000 m, u = 10000 m/s, G = 6.67 × 10-11Nm2kg-2, M = 5.97 × 1024kg

So, h = 2.62 × 107m

This is the height from the surface of the earth upto which the body reaches.

0 is the power due to gravitational force at its highest point

Bohr radius R is given by R = 0.059n²/z nm

=> D = 0.0001 mm = 1×10-⁴ mm = 1000A° => R = 500A°

=> 500 = 0.529n²/1=> n² = 500/0.529=> n = 30.71

so, option D

(i) Mass doesn't change. It remains constant in all situations. so, mass of body at an altitude of 1000km must be 100kg.

(ii) we know, g=g°/(1+h/R)^2

g°=9.8m/s^2

h = 1000km and R = 6400 km

now, g' = 9.8/(1 + 1000/6400)² = 7.33 m/s²

so, weight = mass × acceleration due to gravity

= 100kg × 7.33 m/s² = 733N.

The correct answer is 2No.

Ans :- The gravitational force is a force that attracts any objects with mass. You, right now, are pulling on every other object in the entire universe! This is called Newton's Universal Law of Gravitation.

9.8 m/s2

At the surface of the Earth, the acceleration due to gravity is roughly9.8 m/s2. The average distance to the centre of the Earth is 6371 km. Using the constant k, we can work out gravitational acceleration at a certain altitude.

letGMm = K , a be the distance between M and m .

K/ a^2 = 3K / ( L- a) ^2.

√3a = ( L- a)

a (1 + √3) = L

a = L / ( 1+ √3)

K/ a^2 =K ( 1+ √3)^2/ L^2

= GMm(1+√3)/L^2

In absence of magnetic field, field energy is determined by the principle, quantum number a, while the orbital quantum number. If an electron is innthstate then the magnitude of the angularmomentumis h /2π l(l + 1) where l = 0, 1, 2, ..............,(n - 1),

Since l = 0, 1,,2,:...., (n - 1), different values ofA are compatible with the same value of n. For example, when n = 3, the possible value of lare0,1,2,and when n =4, the possible values of lare0, 1, 2, 3. Thus, the electron in one of the atomscould have n.= 3, I = 2. Therefore, according to quantum mechanics,it is possible forthe electrons to have different energies but havethe same orbital angularmomentum.

by applying formulae m1×v1=m2×v2 we get recoil velocity o.8m/s

Acceleration due to gravity is independent of mass of body.

ascending order ka and

nice question I challenge you I solve question

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ascending order .

Question in correct form is:Two bodies A and B of masses m and 2m respectively are kept a distance d apart. Where a small particle be placed so that the net gravitational force on it due to the bodies A and B is 0.

Answer:Net gravitational force will be zero towards less massive body.Let,The mass of small particle which is being placed is m₀Distance between m₀ and m = xThen, distance between m₀ and 2m = d - x

Force due to m on m₀ = Force due to 2m on m₀Gmm₀/x^2 = G*(2m) * (m₀) / (d - x)^21/√2 = x / (d - x)x√2 = d - xx = d / 2.414x = 0.414 d

The small particle should be kept at a distance 0.414d from A.

The position of center of mass does not change because system is isolated and no external force is acting on the system. Therefore the velocity of the center of mass will be zero.

Let us start from the basics,So,I am assuming the body had a free fall from a height hA and hB respectivleySo,the potential energy stored will be converted into kinetic energySo,(1/2) m v² = mghv=√2ghAnd,t=√(2h/g)So,we conclude that time taken is independent of the massesNow they both reached the ground at the same time so,h = t²g/2So,height and gravity acceleration is directly proportional to each otherSo,hA / hB = gA/ gB

This is the required relation.

The area undet F-t curve , gives the change in momentum of the particle

=> ∆P = A = M.V'

=> V' = A/M

option A.

this can also be done by dimensional analysis..as only A is having the dimension of velocity as asked.

A weighing machine measures the weight of a body and is calibarated to indicate mass. when we stand on a weighing machine our weight acts downwards while upthrust due to air acts upwards. as a result our apparant weight becomes less than the true weight. the weighing machine measures this apparant weight and hence the mass indicated by it is less than actual mass. our actual masa will be more than 42 kg