The time of reverberation of a room A is one second. What will be the time (in seconds) of reverberation of a room,having all the dimensions double of those of room A?

2
4
¹⁄₂
1

Solution

Reverberation time,

The physical quantity having the dimensions[M^-1 L^-3 T³ A²] is

resistance
resistivity
electrical conductivity
electromotive force

Solution

Resistivity,\(\rho =\frac{m}{ne^{2\tau }}\)

ρ=[M L³ T^-3 A^-2]

So, electrical conductivity

σ=¹⁄_ρ

σ=[M^-1 L^-3 T³ A²]

What are the dimensions of permeability ?

[M¹ L¹ T¹ A^-2]
[M¹ L¹ T^-2 A^-2]
[M² L² T¹ A⁰]
[M¹ L² T² A^-2]

Solution

The magnetic field at a point near a long straight conductor is given by

The resistance R of a wire is given by the relation R =\(\frac{\rho \iota }{r^{2}}\).Percentage error in the measurement of r, land r is 1%, 2%and 3% respectively. Then the percentage error in theme asurement of R is :

Solution

What is the fractional error in g calculated from T=2π\(\sqrt{\iota /g}\)? Given fractional errors in T and lare ± x and ± y respectively.

x + y
x– y
2x + y
2x – y

Solution

In an experiment four quantities a, b, c and d are measured with percentage error 1%, 2%, 3% and 4% respectively.Quantity P is calculated as follows \(P=\frac{a^{2}b^{2}}{cd}\) %error in P is

Solution

The density of amaterial in CGS system of units is 4g/cm3.In a system of units in which unit of length is 10 cm and unit of mass is 100 g, the value of density of material will be

0.4
40
400
0.04

Solution

In CGS system,d=\(4\frac{g}{cm^{3}}\)

The unit of mass is 100g and unit of length is 10 cm, so

If the dimensions of a physical quantity are given by M^a L^b T^c, then the physical quantity will be

velocity if a = 1, b =0, c =– 1
acceleration if a= 1, b = 1, c = – 2
force if a = 0, b = – 1, c = – 2
pressure if a= 1, b = – 1, c = – 2

Solution

Pressure=\(\frac{MLT^{-2}}{L^{2}}=[ML^{-1}T^{-2}]\)

⇒a = 1, b = – 1, c = –2.

Two full turns of the circular scale of a screw gauge cover a distance of 1mm on its main scale. The total number of divisions on the circular scale is 50. Further, it is found that the screw gauge has a zero error of – 0.03 mm. While measuring the diameter of a thin wire, a student notes the main scale reading of 3mm and the number of circulr scale divisions in line with the main scale as 35. The diameter of the wire is

3.32 mm
3.73 mm
3.67 mm
3.38 mm

Solution

Least count of screw gauge =\(\frac{0.5}{0.01}\)mm=50mm

∴Reading = [Main scale reading+ circular scale reading× L.C] – (zero error)

= [3 + 35 × 0.01] – (–0.03) = 3.38 mm

The dimensions of magnetic field in M, L, T and C (coulomb)are given as

[M L T^-1 C^-1]
[M T² C^-2]
[M T^-1 C^-1]
[M T^-2 C^-1]

Solution

We know that F = q v B

∴ B=\(\frac{F}{qv}=\frac{MLT^{-2}}{C\times LT^{-1}}=MT^{-1}C^{-1}\)

UnAttempted

0

Attempted

0

Correct

0

Wrong

0

Complete