Physics NCERT Exemplar Solutions Class 12th Chapter Nine

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2 months ago

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P
Payal Gupta

Contributor-Level 10

Maximum Distance = 3 (x + x) = 6x = 3 * 2x = 3 * 50 = 150 cm

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2 months ago

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A
alok kumar singh

Contributor-Level 10

VC=2GMR

Using conservation of Mechanical Energy

GMmR+12*m (Ve29)=GMm (R+h)

1R+h=89Rh=R8=64008=800km

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

siniC=μrarerμdener=velocityofrorervelocityofdenser

sinic=1.5*1082*108=34

tanic=37

ic=tan1 (37)

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2 months ago

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P
Payal Gupta

Contributor-Level 10

Consider the following image dddddddddddddddddddddddddddddddddddddd

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P
Payal Gupta

Contributor-Level 10

r^i^=2cosθn^...... (1)

i^.n^=cosθ...... (2)

r^=i^2 (i^.n^)n^

b=a2 (a.c)c

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Vishal Baghel

Contributor-Level 10

T = 0.5 sec

No. of oscillation = 100

Resolution = 1 sec

l = 1 0 c m ± . 1 c m

               

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2 months ago

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V
Vishal Baghel

Contributor-Level 10

As,  1v1u=1f1v'+f1μ'f=1f

Also,  μ'v'=225

Using Newton's formula - μ'v'=f2

So, f2 = 225

f = 15 cm

 

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2 months ago

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V
Vishal Baghel

Contributor-Level 10

u = 100 cm

f=R2=100cm

after 10 sec i=66=1A

u = 80 m

1v=1f1u=1100180=1801100=54400=1400

v=400cm

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