Units and Measurement

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New answer posted

4 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

 h=50t12? *10t2=80m

Now velocity B with respect to A

VBA=VBVA

= 50 – 30

= 20 m/sec

Now ABA=aBaA

10 (j^) (10j^)

= 0

Time taken to collide.

h=VBAt

80 = 20t

t = 4 sec

Total time = 4 + 2

= 6 sec

New answer posted

4 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Least count = 0.550mm = 0.01 mm

Diameter, d = 1.5 + 7 × 0.01

= 1.57

 Surface Area = (2r) l

= 3.4 cm2

New answer posted

4 months ago

0 Follower 6 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

4 months ago

0 Follower 5 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

4 months ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

MS – Reading = 2.5 mm.

50 division on CS = 0.5 mm on MS.

45thdivisiononCS=0.550*45mmM.S.

= .45 mm on M.S.

So, diameter = 2.5 mm

+0.45 mm

+0.03 mm

= 2.98 mm

New answer posted

4 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

MSD = 20 divisions per cm 1 MSD = 120cm.

VSD = 50 divisions

As, 25 VSD = 24 MSD

1VSD=2425MSD

L.C. = 1 MSD – 1 VSD

=1500cm=0.002cm

New answer posted

4 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

 T=kρr3S3/2

[T]  [ML3]1/2 [L3]1/2 [MLT2L1]34

[T] = [M1/2L32L32M34T32]

= [M14T3/2]

New answer posted

4 months ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

T=kρr3S3/2

[T]  [ML3]1/2 [L3]1/2 [MLT2L1]34

[T] = [M1/2L32L32M34T32]

= [M14T3/2]

New answer posted

4 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Sol.

BA=μ0Iθ4πR

BABB=IAθARBIBθBRA

23π2 (4)35π3 [2]

65 .

New answer posted

4 months ago

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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