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we know acceleration is the change in velocity with time

therefore, from this statement we can say that object p has no velocity because it is not changing its velocity with time , it is moving with a constant velocity , therefore it's acceleration will be 0

but we can say that object q has acceleration because it is said to changing the velocity with time as written in the questions

w = 2pi*2/pi = 4 rad/s.

npw consider angle thethaa wth vertical and do comonents of Tension along horizontal and vertial and balnce them with mg and mw^2r

where from geomtery r = lsinthetha.

after balancing the eqns would be:

T = mw^2l

T = 100/1000 * 16 * 1 = 1.6 N

in vertical direcn.

Tcosthetha = mg

cosththa = 5/18

or

thetha = arccos(5/18)

Let the required tension be T and the tension in the string connecting A and B be T1

equations are:

for C = 2 g - T = 2 a (i)

for B = 2 g + T - T1= 2 a (ii)

and T1- 2 g = 2 a (iii)

adding (i), (ii), (iii)

2 g = 6 a or a = g/3

substituting in (i) = T = 2(g -a) = 2(g - g/3) = 4 g/3

= 4(9.81)/3 = 13.08 N

Let two equal charges Q placed at point A and B joining line segment. Let length of line segment is x and charge q is placed at centre.

Then distance of charge q from point A and B is x/2

Force on the charge at B due to the charge at A is,

F1 = kQ2^2 /x^2

Force on the charge at B due to the charge q is,

F2 = kQq/(x/2)^2

For the system to be in equilibrium,

F1 + F2 = 0

=> kQ^2/x^2 + kQq/(x/2)^2 = 0

=> q = -Q/4

Option D

c is the best answer

answer of the question is C

Let Q1 be a test positive charge, since charges of Q2 and Q3 are opposite in sign, there exists an attractive force between them. One point to be kept in mind here is that the net force on Q3 must be 0 as given in the problem. In order to keep force on Q3 be 0, the force between Q1 and Q3 must be force of repulsion as we already have attractive force between Q2 and Q3Now to satisfy the equilibrium position, the two forces must balance each other.KQ1Q3/4R2 = -KQ2Q3/R2….. (distance between Q1 and Q3 is 2R).Q1 = -4Q2Hence charge of Q1 is opposite to that of Q2 and same as that of Q3

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Using symmetry here we see that that all froces

=) F1 = F2 = F3 = kQq/r^2

here r = AO = BO = CO

using trigonometry we can find " r "

Here we can see that length of each side of ∆ ABC is " L "

=) In ∆ BOD

Angle OBD = 30°

and BD = L/2

COS 30 ° = L/2r =) r = L/2 × sec30°

=). r = L/2 × 2/√3 = ( 1/√3 ) L -----(1)

so we can see tha all forces are equal

F1 = F2 = F3 = kQq/ (L/√3)^2 = 3kQq/L^2

and angle between each forces is 120°

using Lami's theorem if all forces are equal in magnitude and angle between then is also equal and planar then resultant force is 0

=) Fnet = F1 +F2 + F3 = 0

=) here we add forces using vectors to get net force = 0

so Net force on charge " Q " placed at centroid is zero.

not satisfactry

net force acting on the system is m2*g = 8*10 = 80

and total mass of the system is (8+12) = 20kg

so, acceleration of the system = 80/20 = 4m/s²

net force acting on the system is m2*g = 8*10 = 80

and total mass of the system is (8+12) = 20kg

so, acceleration of the system = 80/20 = 4m/s²

the focal length of the convex mirror is 80

thank u

Potential at any point inside the conductor will be same as surfaceSo V =K /rWhere k is constant and r is the distance from centreSo V =k/ 10K =10VVat r=15=k/15=10V/15=2/3V

As the object moves towards the focus of the lens the size of the image increases and it moves away from the focus. ... When the object is at a distance less than the focal length, the image formed is virtual, enlarged and erect. It is formed on the same side of the lens as that of the object.

To get a real image of same size , the object should placed at a distance 2f from the optical center in the front of the lens

The object should be placed between 15cm and 30 cm( Between f and 2f) from the converging lens to obtain a real and magnified image.

a)

b)

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The following are the differences between a real image and virtual image:

A real image is formed when the rays of light after reflection or refraction actually meet at some point.

A virtual image is formed when the rays of light after reflection or refraction appear to meet at a point.

answer of is question is c d

Magnification:-~~~~~~~~~~~~~~☆Magnification produced by a spherical mirror or lens can be defined as the ratio of the height of the image to that of the object.

☆If we take h as the height of the object and h' as the height of the image, then the Magnification produced by the mirror or lens will be:-

Magnification, m = h'/h

☆The sign of the Magnification for real image is taken to be as (-)ve and it is taken as (+)ve for virtual images.

true all answer.

TrueTrueFalseTrueTrueFalse

Given, magnification (m) = -3 (image is real)object-distance (u) = -10 cmimage-distance (v) = ?Now, In a mirror m= -v/u = -3 = -v/-10 -3 x 10 = v v = -30 cmSo, the object is placed 30 cm in front of the mirror.

Visual tasks such as reading, spending hours at the computer or watching television require you to focus and concentrate – and without you realizing it, you are actually placing strain on your eye muscles. Theeye muscles tightenand cause your eyes to become dry, irritated and uncomfortable..

Musicissounds organized in time. By the way, with regard to the physics ofsound. Noise refers to individualsoundsthat do not have a regular oscillating waveform and they have no recognizable pitch.

All the time the noise is above 65db but music (without home theater) is below the 65db.

music is bearable noise is unbearable

moving bike possess kinetic energy.

while climbing the stairs, we store potential energy

Table fan, electric fan,

100/10 m/s^2=10 m/s^

10 is the correct answer of the given question