Conversion of 1 MW power in a New system of units having basic units of mass, length and time as 10 kg, 1 dm and 1minute respectively is
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Solution
We have, n1u1 = n2u2
n2=n2\(\frac{u_{1}}{u_{2}}\)
=106×\(\left (\frac{M_{1}}{M_{2}} \right )\left ( \frac{L_{1}}{L_{2}} \right )^{2}\left ( \frac{T_{1}}{T_{2}} \right )^{-3}\)
=106×\(\left ( \frac{1kg}{10kg} \right )\left ( \frac{1m}{1\times 10^{-1}m} \right )^{2}\left ( \frac{1s}{60s} \right )^{-3}\)
=106×\(\left ( \frac{1}{10} \right )(10)^{2}(60)^{3}\)
=107×(60)3= 2.16 × 1012 units.
What is the fractional error in g calculated from T=2π\(\sqrt{\iota /g}\)? Given fraction errors in T and l are ± x and ± y respectively.
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Solution
From T=2π\(\sqrt{\frac{\iota }{g}}\) ; g\(4\, \pi ^{2}\frac{\iota }{T^{2}}\)
\(\frac{\Delta g}{g}=\frac{\Delta \iota }{\iota }+\frac{2\Delta T}{T}=(y+2x)\)
In a Vernier calliper, Ndivisions of vernier scale coincide with(N–1) divisions of main scale (in which one division represents 1 mm). the least count of the instrument in cm.should be
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Solution
N VD = (N– 1) MD
1 VD=\(\left ( \frac{N-1}{N} \right )\)MD
L.C. = Least count = 1MD – 1VD
L.C.=\(\left ( 1-\frac{N-1}{N} \right )\)MD
=\(\left ( 1-\frac{N-1}{N} \right )\)
=\(\frac{value\, of\, 1\, part\, on\, main\, scale}{number\, of\, parts\, on\, vernier\, scale}\)
where V.D. = vernier division, M.D. Main scale division.
A quantity is represented by X = Ma Lb Tc. The % error in measurement of M,L and T are α%, β% and γ% respectively. The % error in X would be
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Solution
X=Ma Lb Tc ;
\(\frac{\Delta X}{X}\times 100=\left ( \frac{a\, \Delta M}{M}+\frac{b\, \Delta L}{L}+\frac{c\, \Delta T}{T} \right )\times 100\)
=(α a + β b + γ c)%
A spherical body of mass m and radius r is allowed to fall in a medium of viscosity η. The time in which the velocity of the body increases from zero to 0.63 times the terminal velocity (v) is called time constant τ. Dimensionally τ can be represented by
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Solution
None of the expressions has the dimensions of time.
A cube has numerically equal volume and surface area. The volume of such a cube is
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Solution
Volume (L3) = surface area (6L2)
∴L =6,volume = 63 = 216
Which of the following do not have the same dimensional formula as the velocity?Given that μ0= permeability of free space, ε0= permittivity of free space, v= frequency, λ= wavelength,P = pressure, ρ= density, ω= angular frequency, k = wave number
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Solution
ωK=\(\frac{1}{T}\times \frac{1}{L}=[L^{-1}T^{-1}]\)
The dimensions of the quantities in a, b,c are of velocity[LT-1]
The mass of the liquid flowing per second per unit area of cross-section of the tube is proportional to (pressure difference across the ends)n and (average velocity of the liquid)m. Which of the following relations between m and n is correct?
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Solution
Let M = pnvm
ML-2T-1=(ML-1T-2)n(LT-1)m
∴ n = 1;-n + m =- 2
∴ m = -2+n = -2 + 1 =-1 ∴ m=-n
The Richardson equation is given by I= AT2e–B/kT. The dimensional formula for AB2 is same as that for
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Solution
I = AT2 e–B/kT
Dimensions of A = I /T2;Dimensions of B = kT
(∵power of exponential is dimensionless)
AB2=\(\frac{I}{T^{2}}\)(KT)2=IK2
Dimensions of‘ohm’ are same as (where h is Planck’s constant and e is charge)
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Solution
\(\frac{h}{e^{2}}=\frac{ML^{2}T^{-1}}{(AT)^{2}}=ML^{2}T^{-3}A^{-2}\)= Resistance (ohm)