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a year ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Twelve

For a perfect gas, two pressures P1 and P2 are shown in figure. The graph shows:
0 Follower 27 Views

P
Payal Gupta

Contributor-Level 10

At constant Temp

P α 1 v

v2 > v1

⇒ P1 > P2

New answer posted

a year ago

Physics Gravitation

Four spheres each of mass m form a square of side d (as shown in figure). A fifth sphere of mass M is situated at the centre of square. The total gravitational potential energy of the system is
0 Follower 14 Views

P
Payal Gupta

Contributor-Level 10

Amount of work done to bring mass from infinity to point (1) u1 = 0

Similarly, to bring mass (2) u 2 = − G m d m

u 3 = − G m d [ m 1 + m 2 ]

u 4 = − G m d [ 2 m + m 2 ]

u 5 = − G m d [ 4 2 M ]

U T = u 1 + u 2 + u 3 + u 4 + u 5

U T = − G m d [ ( 4 + 2 ) m + 4 2 M ]

             
     

New answer posted

a year ago

Physics System of Particles and Rotational Motion

A stone tide to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of change in its velocity, as it reaches a position where the string is horizontal, is x ( h 2 − g L ) .  The value of x is
0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

Conservation of Energy b/w (1) & (2)

1 2 m v 2 + m g l = 1 2 m u 2

v = u 2 − 2 g l

Δ v → = v j ^ − u i ^

| Δ v → | = ( u ) 2 + ( u 2 − 2 g l ) 2

= u 2 + u 2 − 2 g l

= 2 ( u 2 − g l )

x = 2

New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Five

One end of a massless spring of spring constant k and natural length l 0  is fixed while the other end is connected to a small object of mass m lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity ω  about an axis passing through fixed end, then the elongation of the spring will be
0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

F.B.D, with respect to Non-Inertial frame

⇒ k x = m ω 2 ( l 0 + x )

⇒ x | k − m ω 2 ( l 0 + x )

x = m ω 2 l 0 k − m ω 2

        

               

New answer posted

a year ago

Physics Motion in a Straight Line

When a ball is dropped into a lake form a height 4.9 m above the water level, it hits the water with a velocity v and then sinks to the bottom with the constant velocity v. It reaches the bottom of the lake 4.0 s after it is dropped. The approximate depth of the lake is
0 Follower 16 Views

P
Payal Gupta

Contributor-Level 10

Total time (T) = 4 sec

Given u = 0 m/sec

a = g = 9.8 m/sec2

h = 4.9 m, t =?

⇒ h = u t + 1 2 a t 2                

4 . 9 = 0 . t + 1 2 * 9 . 8 t 2                

t = 1 sec

'v' be the velocity with which ball hits the water v = u + at

= 0 + 9.8 * 1 = 9.8 m/sec

Time taken to reach the bottom of the lake from surface of the lake

= 4 – 1 = 3 sec

v = 9.8 m/sec

H = u t + 1 2 g t 2 = 9 . 8 * 3 + 1 2 * 9 . 8 * 9                

29.4 + 4.9 * 9 = 29.4 + 44.1

H = 73.5 m

New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

The distance of the Sun from earth is 1.5 * 1011 m and its angular diameter is (2000) s when observed from the earth. The diameter of the Sun will be
0 Follower 21 Views

P
Payal Gupta

Contributor-Level 10

1° →  60'

⇒ 60' → 10°

6 0 ' → π 1 8 0 C

θ = π * 2 * 1 0 3 1 8 0 * 3 6 0 0          

L = 1.5 * 1011 m

D = L θ = 1 . 5 * 1 0 1 1 * π 1 8 0 * 3 6 0 0 * 2 * 1 0 3 = 1 . 4 5 * 1 0 9 m    

   

New question posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 12th Chapter Nine

The distance of the Sun from earth is 1.5 × 1011 m and its angular diameter is (2000) s when observed from the earth. The diameter of the Sun will be
0 Follower 2 Views

New question posted

a year ago

Physics Ray Optics and Optical Instruments

The distance of the Sun from earth is 1.5 × 1011 m and its angular diameter is (2000) s when observed from the earth. The diameter of the Sun will be:
0 Follower 1 View

New answer posted

a year ago

Physics physics ncert exemplar solutions class 12th chapter two

The SI unit of a physical quantity is pascal-second. The dimensional formula of this quantity will be:
0 Follower 10 Views

P
Payal Gupta

Contributor-Level 10

S. I. unit Pascal – second   = N m 2 − s e c  

= M L T − 2 T L 2  

= M L − 1 T − 1  

New answer posted

a year ago

Physics Units and Measurement

The distance of the Sun from earth is 1.5 * 1011 m and its angular diameter is (2000) s when observed from the earth. The diameter of the Sun will be
0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

1°  ® 60'

              Þ 60' ® 10°

               

               

              L = 1.5 * 1011 m

               

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