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Physics 2 Marks Questions

Introduction to Physics · Uttarakhand Board (UBSE) STD 12 Science (Uttarakhand - English Medium). 16 questions.

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Question 12 Marks
The metre is defined as the distance travelled by light in $\frac{1}{299,792,458}$ second. Why didn't people choose some 1 easier number such as $\frac{1}{300,000,000}$ second? Why not 1 second?
Answer
Until 1960, meter was defined as the distance between two lines on a specific bar of platinum iridium alloy stored under controlled condition. Because of its limited accuracy, it was abandoned by scientist. From the above scale speed of light was 299792458m/ s and not $300000000\ ms^{-1}$. They needed something which could give them more accurate results in different places and at different times. Thus, in 1983, they adopted distance travelled by light in $\frac{1}{\text{c}}\text{s}.$
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Question 22 Marks
Find the dimensions of:
  1. The specific heat capacity c.
  2. The coefficient of linear expansion a.
  3. The gas constant R.
Some of the equations involving these quantities are$\text{Q = mc(T}_2-\text{T}_1),\text{l}_{\text{t}}=\text{l}_0[1+\alpha(\text{T}_2-\text{T}_1)]$ and $\text{PV = nRT}.$
Answer
  1. Specific heat capacity $=\text{C}=\frac{\text{Q}}{\text{m}\triangle\text{T}}=\frac{\text{ML}^2\text{T}^{-2}}{[\text{M}][\text{k}]}=[\text{L}^2\text{T}^{-2}\text{K}^{-1}]$
  2. Coefficient of linear expansion $=\alpha=\frac{\text{L}_{1}-\text{L}_{2}}{\text{L}_0\triangle\text{T}}=\frac{[\text{L}]}{[\text{L}][\text{R}]}=[\text{k}^{-1}]$
  3. Gas constant $=\text{R}=\frac{\text{PV}}{\text{nT}}=\frac{[\text{ML}^{-1}\text{T}^{-2}][\text{L}^3]}{[(\text{mol})][\text{k}]}=[\text{ML}^2\text{T}^{-2}\text{k}^{-1}(\text{mol})^{-1}]$
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Question 32 Marks
Find the dimensions of:
  1. Electric field E,
  2. Magnetic field B and
  3. Magnetic permeability $\mu_0.$
The relevant equations are$\text{F = qE, F = qvB,}$ and $\text{B}=\frac{\mu_0\text{I}}{2\pi\alpha};$
where F is force, q is charge, v is speed, I is current, and a is distance.
Answer
  1. Electric field $\text{E}=\frac{\text{F}}{\text{q}}=\frac{\text{MLT}^{-2}}{[\text{lT}]}=[\text{MLT}^{-3}\text{l}^{-2}]$
  2. Magnetic field $\text{B}=\frac{\text{F}}{\text{qv}}=\frac{\text{MLT}^{-2}}{[\text{lT}][\text{LT}^{-1}]}=[\text{MT}^{-2}\text{l}^{-1}]$
  3. Magnetic permeability $\mu_0=\frac{\text{B}\times2\pi\text{a}}{\text{l}}=\frac{[\text{MT}^{-2}\text{l}^{-1}]\times[\text{L}]}{[\text{l}]}=[\text{MLT}^{-2}\text{l}^{-2}]$
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Question 42 Marks
It is desirable that the standards of units be easily available, invariable, indestructible and easily reproducible. If we use foot of a person as a standard unit of length, which of the above features are present and which are not?
Answer
Foot-easily available. It is not invariable. It is not indestructible. It is not easily reproducible, the length may change in time over years, as people grow and get diseases which can alter the length of foot.
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Question 52 Marks
Suggest a way to measure:
  1. The thickness of a sheet of paper.
  2. The distance between the sun and the moon.
Answer
  1. By intensity of light from sun directly and that of light reflected by Moon.
  2. At the time of eclipse, we can find the distance between Earth and Sun and distance between Earth and Moon. Then we subtract or add to get the distance.
Using the lunar year, Earth's year and distance between earth and moon.
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Question 62 Marks
Let I = current through a conductor, R = its resistance and V = potential difference across its ends. According to Ohm's law, product of two of these quantities equals the third. Obtain Ohm's law from dimensional analysis. Dimensional formulae for R and V are $\text{ML}^2\text{I}^{-2}\text{T}^{-3}$ and $\text{ML}^2\text{T}^{-3}\text{I}^{-1}$ respectively.
Answer
Dimensional formulae of $\text{R}=[\text{ML}^2\text{T}^{-3}\text{I}^{-2}]$ Dimensional formulae of $\text{V}=[\text{ML}^{2}\text{T}^3\text{I}^{-1}]$ Dimensional formulae of $\text{I}=[\text{I}]$$\therefore[\text{ML}^2\text{T}^3\text{I}^{-1}]=[\text{ML}^2\text{T}^{-3}\text{I}^{-2}][\text{I}]$
$\Rightarrow\text{V = IR}$
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Question 72 Marks
Suppose you are told that the linear size of everything in the universe has been doubled overnight. Can you test this statement by measuring sizes with a metre stick? Can you test it by using the fact that the speed of light is a universal constant and has not changed? What will happen if all the clocks in the universe also start running at half the speed?
Answer
No, Yes, No After clock becomes slow, when 1 hr passes in clock, 2 hr passes in reality. l = ct : old clock time 2l = c (2t) : new clock time$\therefore$ l = ct.
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Question 82 Marks
Find the dimensions of:
  1. Linear momentum.
  2. Frequency.
  3. Pressure.
Answer
  1. Linear momentum: $\text{mv}=[\text{MLT}^{-1}]$
  2. Frequency: $\frac{1}{\text{T}}=[\text{M}^0\text{L}^0\text{T}^{-1}]$
  3. Pressure: $\frac{\text{Force}}{\text{Area}}=\frac{[\text{MLT}^{-2}]}{\text{[L}^2]}=[\text{ML}^{-1}\text{T}^{-2}]$
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Question 92 Marks
The height of mercury column in a barometer in a Calcutta laboratory was recorded to be 75cm. Calculate this pressure in SI and CGS units using the following data: Specific gravity of mercury $=13.6$, Density of water $=10^3 \mathrm{~kg} / \mathrm{m}^3, \mathrm{~g}=9.8 \mathrm{~m} / \mathrm{s}^2$ at Calcutta. Pressure = $\text{h}\rho\text{g}$ in usual symbols.
Answer
Height $\mathrm{h}=75 \mathrm{~cm}$, Density of mercury $=13600 \mathrm{~kg} / \mathrm{m}^3, \mathrm{~g}=9.8 \mathrm{~ms}^{-2}$ then
Pressure $=\mathrm{hfg}=10 \times 10^4 \mathrm{~N} / \mathrm{m}^2$ (approximately)
In C.G.S. Units, $P=10 \times 10^5$ dyne $/ \mathrm{cm}^2$
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Question 102 Marks
Taking force, length and time to be the fundamental quantities find the dimensions of:
  1. Density.
  2. Pressure.
  3. Momentum.
  4. Energy.
Answer
Taking force, length and time as fundamental quantity:
  1. $\text{Density}=\frac{\text{m}}{\text{V}}=\frac{\big(\frac{\text{force}}{\text{acceleration}}\big)}{\text{Volume}}\\=\frac{\big[\frac{\text{F}}{\text{LT}}^{-2}\big]}{[\text{L}^{2}]}=\frac{\text{F}}{\text{L}^4\text{T}^{-2}}=[\text{FL}^{-4}\text{T}^{2}]$
  2. $\text{Pressure}=\frac{\text{F}}{\text{A}}=\frac{\text{F}}{\text{L}^2}=[\text{FL}^{-2}]$
  3. $\text{Momentum = mv (Force I acceleration)}\times\text{Velocity}\\=\Big[\frac{\text{F}}{\text{LT}^{-2}}\Big]\times[\text{LT}^{-1}]=[\text{FT}]$
  4. $\text{Energy}=\frac{1}{2}\text{mv}^2=\frac{\text{Force}}{\text{acceleration}}\times\text{(velocty)}^2$
$=\Big[\frac{\text{F}}{\text{LT}^{-2}}\Big]\times[\text{LT}^{-1}]^2=\Big[\frac{\text{F}}{\text{LT}^{-2}}\Big]\times[\text{L}^2\text{T}^{-2}]=[\text{FL}]$
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Question 112 Marks
Find the dimensions of:
  1. Electric dipole moment p.
  2. Magnetic dipole moment M.
The defining equations are p = q.d and M = IA, where d is distance, A is area, q is charge and I is current.
Answer
  1. Electric dipole moment $\text{P = qI = [IT]}\times[\text{L] = [LTI}]$
  2. Magnetic dipole moment $\text{M = IA}=[\text{I][L}^2]=[\text{L}^2\text{I}]$
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Question 122 Marks
Find the dimensions of:
  1. Angular speed $\omega.$
  2. Angular acceleration $\alpha.$
  3. Torque $\tau$ and
  4. Moment of interia I.
Some of the equations involving these quantities are$\omega=\frac{\theta_2-\theta_1}{\text{t}_2-\text{t}_1},\alpha=\frac{\omega_2-\omega_1}{\text{t}_2-\text{t}_1},\tau=\text{F.r}$ and $\text{I = mr}^2.$
The symbols have standard meanings.
Answer
  1. Angular speed $\omega=\frac{\theta}{\text{t}}=[\text{M}^0\text{L}^0\text{T}^{-1}]$
  2. Angular acceleration $\alpha=\frac{\omega}{\text{t}}=\frac{\text{M}^0\text{L}^0\text{T}^{-1}}{\text{T}}=[\text{M}^0\text{L}^0\text{T}^{-2}]$
  3. Torque $\tau=\text{Fr}=[\text{MLT}^{-2}][\text{L] = [ML}^2\text{T}^{-2}]$
  4. Moment of inertia $=\text{Mr}^2=[\text{M}][\text{L}^2]=[\text{ML}^2\text{T}^0]$
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Question 132 Marks
The average speed of a snail is 0.020 miles/hour and that of a leopard is 70 miles/hour. Convert these speeds in SI units.
Answer
The average speed of a snail is 0.02mile/hr
Converting to S.I. uniits, $\frac{0.02\times1.6\times1000}{3600}$ m/sec $[1$mile $= 1.6km = 1600m] = 0.0089ms^{-1}$
The average speed of leopard= 70 miles/hr
In SI units = 70 miles/hour $=\frac{70\times1.6\times1000}{3600}=31\text{m}/\text{s}$
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Question 142 Marks
If two quantities have same dimensions, do they represent same physical content?
Answer
Two physical quantities have same dimensions, but they can have different units. 5 meters, 5cm, 5mm all have dimension as Length. [L]. But different units and different magnitude values.
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Question 152 Marks
Suppose the acceleration due to gravity at a place is $10m/s^2$. Find its value in cm/(minute)$^2$.
Answer
$\text{g}=10\frac{\text{metre}}{\text{sec}^2}=36\times10^5\text{cm/min}^2$
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Question 162 Marks
Find the dimensions of Planck's constant h from the equation E = hv where E is the energy and v is the frequency.
Answer
E = hv, where E = energy and v = frequency.$\text{h}=\frac{\text{E}}{\text{v}}=\frac{[\text{ML}^2\text{T}^{-2}]}{[\text{T}^{-1}]}[\text{ML}^2\text{T}^{-1}]$
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