Class 12th

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New answer posted

a year ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

 A=Aoe−λt

⇒2250=4250e−λt

⇒λ*10*0.434=0.274

⇒λ=0.27410*0.434=0.63min−1

New answer posted

a year ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

U = 3PV + 4

⇒n CvT=3PV+4

⇒n*fRT2=3PV+4

⇒f=6+8PV

⇒f>6⇒poly  atomic

New answer posted

a year ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

 λ1=6MHz=6*106Hz

λ1=Cυ1=3*1086*106=50m

λ2=Cυ2=10*106=3*10810*106=30m

Wavelength bandwidth

=|λ1−λ2|=20m

New answer posted

a year ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

λ = c v = Q 3 W 2 = I 3 T 3 M 2 L 4 T − 4 ⇒ [ λ ] = [ M − 2 L − 4 T 7 l 3 ]

New answer posted

a year ago

0 Follower 22 Views

V
Vishal Baghel

Contributor-Level 10

Given Acceleration of m2 is 2m/s2

(upward)

N = 20a              T – m2g = m2a

T – 20g = 20a

T = 200 + 20 * 2

T = 240 N

T = m1 a1

240 = 10a1

a1 = 24 m/s2

⇒ T a + T a 1 − T a 2 = 0        

a 2 = 2 4 + 2 = 2 6 m / s 2                        

F – T – N = Ma2

F – 240 = 100 * 26 + 20 * 26

F = 3360 N

New answer posted

a year ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

F=ΔρΔt=1.8− (−1.8)Δt=100

⇒Δt=3.6100=0.036  sec

New answer posted

a year ago

0 Follower 8 Views

P
Payal Gupta

Contributor-Level 10

tanα=vmaxt1=a1

tanβ=vmaxt2=a2

⇒a1a2=t2t1⇒t1t2=a2a1

New answer posted

a year ago

0 Follower 8 Views

A
alok kumar singh

Contributor-Level 10

1 MSD = 1mm

9 MSD = 10 VSD

1VSD = 0.9 MSD = 0.9 mm

Least count = 1 MSD – 1 VSD

= 0.1 mm = 0.01 cm

Zero Error = (10 – 4) * 0.1 = 0.6 mm

Reading = MSR + VSR – Zero error

= 3 cm + 6 * 0.01 – (0.06)

= 3.12 cm ?  3.10 cm

New answer posted

a year ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

v=ωA2−x2⇒v2A2ω2+x2A2=1 Path is ellipse.

New answer posted

a year ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

R1=u2sin90g

R2=u2sin60g

⇒R1R2=23

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