Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

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New answer posted

2 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

 mN2=14g, T=27°C=300k

To double the Vrms Tf = 4Ti = 4 * 300 = 1200 k.

Q=nCvΔT

=5*225R=1125R=1125*3.32

= 9360 J.

New answer posted

2 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

ρω=1000kg/m3

G = 10 m/s2

Using bernaulli's eqn

5*105π+ρg*10=ρ0+12ρVe2

=318? 17.8m/s

New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

 T1T2=46=g2g1

(1+hR)2=941+hR=32h=R2=3200km.

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

m = 200 g, ki = 90 J kf = 40 J.,  Δt=1sec

F=kfkix.4*90.4*40=F*1

F=64=2N

So for complete rest :-

Fx = 90 J

2x = 90 x = 45 m

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

Mass = m

Length = L

mL=?

at l=Lx, achain=g2

a=FPullingmass= (L? l)? g? ? lg? L= (L? 2l)? g? L=g2

l=L4? x=4

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

A = 10 cm2, V = 20 m/s

F=dpdt=ρAV2=103*10*104*400

= 400 N.

New answer posted

2 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

x = 3t  y = 5t3

Vx=dxdt=3Vy=dydt=15t2

ax=dVxdt=0ay=dVydt=30t

a=30tj^

at t = 1 sec,  a=30j^

New answer posted

2 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

Efficiency η=αBsinθlogeBxkTα, Bconstants

As, energy =12kT Dimension of Bx = Dimension of energy

So Dimension as of B = Dimension of 'F'

Dimension of F1

Dimension of  (1α)x= Dimension of energy

New answer posted

2 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

m (L)=m1S1 (ΔT)

m3.4*105= (200) (4200) (25)

m=61.7

New question posted

2 months ago

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