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New answer posted

a year ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

V r m s = 3 R T M A s .         v 1 = v 2

V r m s α T       η 2 = 4 n 1

 

T 1 = T 2 ⇒ V r m s − 1 = V r m s − 2

P 1 = n 1 R T v P z = n 2 R T v

P 1 P 2 = n 1 n 2 = 1 4

 

 

New answer posted

a year ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

A.A is correct statement

η = 1 − T 2 T 1 = 1 − 1 0 0 4 0 0 = 3 4 = 7 5 %

 

New answer posted

a year ago

0 Follower 12 Views

A
alok kumar singh

Contributor-Level 10

K 2 k 1 = 9 A 1 A 2 = 2 , L 1 L 2 = 2

4 5 0 − T L 1 A 1 k 1 = T L 2 k 2 A 2 ⇒ ( 4 5 0 − T ) L 2 K 2 A 2 = ( T ) L 1 A 1 k 1

( 4 5 0 − T ) = T ( L 1 L 2 k 2 A 2 k 1 A 1 ) ⇒ 4 5 0 − T = T ( 2 * 9 * 1 2 ) = 9 T

10 T = 450

T = 45° C

New answer posted

a year ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

According to definition of wave number, we can write

1λ=R (112−1n2)⇒1−1n2=1λR

⇒ 1 n 2 = 1 − 1 λ R = λ R − 1 λ R ⇒ n = λ R λ R − 1

New answer posted

a year ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

 

E A E B = − G M m A 2 * 3 r − G M m B 2 * 4 r = m A m B * 4 3 = 4 3 * 4 3 = 1 6 9

New answer posted

a year ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

H=l2Rt

%error  in  H=ΔHH*100=2 (Δll)*100+ (ΔRR)*100+ (Δtt)*100=2*2+1+3=8%

New answer posted

a year ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

H=u2sin2θ2g  and  R=u2sin2θg

According to question

R=H⇒u2sin2θ2g=2u2snθcosθg⇒tanθ=4

New answer posted

a year ago

0 Follower 22 Views

A
alok kumar singh

Contributor-Level 10

ρ = 1 0 3 k g / m 3

g = 10 m/s2

Final height in both vessels

= 1 0 0 + 1 5 0 2 = 1 2 5 c m

So, less in U = ∫ A * 0 . 2 5 * g * 0 . 2 5

1 0 3 * 6 2 5 * 1 0 − 4 * 1 0 * 1 6 * 1 0 − 4

= 625 * 16 * 104

= 1J.

New answer posted

a year ago

0 Follower 6 Views

P
Payal Gupta

Contributor-Level 10

v=mv0M+m⇒Kf=12 (M+m)v2=mv022 (M+m)

Ki=mv022

Loss in Kinetic energy = ΔK=Ki−Kf

ΔK=12mv02 (MM+m)=12*0.2*100= [9.810]=9.8J

New answer posted

a year ago

0 Follower 8 Views

P
Payal Gupta

Contributor-Level 10

As we know that gP=g (RR+h)2=g (R54R)2=16g25

⇒Δg=g−gP=9g25

The percentage decrease in the weight of the object = Δgg*100=36%

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