Multiply 107.88 by 0.610 and express the result with correct number of significant figures.

65.8068
65.807
65.81
65.8

Solution

Number of significant figures in multiplication is three,corresponding to the minimum number 107.88 × 0.610 = 65.8068 = 65.8

Subtract 0.2J from 7.26 J and express the result with correct number of significant figures.

7.1 J
7.06 J
7.0 J
7 J

Solution

Subtraction is correct upto one place of decimal,corresponding to the least number of decimal places.7.26 – 0.2 = 7.06 = 7.1 J.

Error in the measurement of radius of a sphere is 1%. Then error in the measurement of volume is

Solution

V=⁴⁄₃πr³;

\(\frac{\Delta V}{V}\times 100=3\left ( \frac{\Delta r}{r} \right )\times 100\)=3×1%=3%

The dimensions of Hubble’s constant are

[T^-1]
[M⁰ L⁰ T^-2]
[MLT⁴]
[MT^-1]

Solution

Hubble’s constant,H=\(\frac{velocity}{distance}=\frac{[LT^{-1}]}{[L]}\)

=[T^-1]=70×10^-3N / m.

The velocity v of a particle at time t is given by v=at+\(at+\frac{b}{t+c}\) The dimensions of a, b c are respectively

[LT^-2], [L], [T]
[L²], [T] and [LT²]
[LT²], [LT] and [L]
[L], [LT] and [T²]

Solution

As c is added to t, ∴c = [T]

a =\(\frac{v}{t}=\frac{LT^{-1}}{T}=[LT^{-2}]\),

b=v(t+c)=LT^-1×T=[L]

The velocity of a body which falls under gravity varies as g^a h^b, where g is acc. due to gravity and h is the height. The values of a and b are

a =1, b =1/2
a= b = 1
a= 1/2, b = 1
a= 1/2; b =1/2

Solution

v=g^ah^b; [M⁰LT^-1]=(LT^-2)^aL^b=L^a+bT^-ab

∴ a + b = 1 ; -2a =-1 a = 1/2

∴ b= 1/2

The deBroglie wavelength associated with a particle of massm and energy E is h/\(\sqrt{2mE}\). The dimensional formula of Planck’s constant h is

[M² L² T^-2]
[ML² T^-1]
[MLT^-2]
[ML²T^-2]

Solution

h=\(\lambda \sqrt{2mE}=L\sqrt{M(ML^{2})}=[ML^{2}T^{-1}]\)

Distance travelled by a particle at any instant ‘t’ can be represented as S = A (t + B) + Ct². The dimensions of B are

[M⁰L¹T^-1]
[M⁰L⁰T¹]
[M⁰L^-1T^-2]
[M⁰L²T^-2]

Solution

In S = A (t + B) + Ct²; B is added to time t. Therefore,dimensions of B are those of time.

The potential energy of a particle varies with distance x from a fixed origin as V=\(\frac{A\sqrt{X}}{X+B}\) where A and B are constants.The dimensions of AB are

[M¹L^5/2T^-2]
[M¹L²T^-2]
[M^3/2L^5/2T^-2]
[M¹L^7/2T^-2]

Solution

B = x = [L]; A=\(\sqrt{x}\)=Vx; A=V\(\sqrt{x}\)

= ML²T^-2L^1/2=ML^5/2T^-2

AB=(ML^5/2T^-2) (L)=[M¹L^7/2T^-2]

The time of oscillation T of asmall drop of liquid depends on radius r, density ρ and surface tension S. The relation between them is given by

\(T\alpha \sqrt{\frac{s}{\rho r^{3}}}\)
\(T\alpha \sqrt{\frac{\rho r^{3}}{s}}\)
\(T\alpha \sqrt{\frac{s^{2}r^{3}}{\rho }}\)
\(T\alpha \sqrt{\frac{\rho r^{3}}{s }}\)

Solution

\(\sqrt{\frac{\rho r^{3}}{S}}=\sqrt{\frac{ML^{-3}L^{3}}{MT^{-2}}}=T\)

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