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Physics 2 Marks Questions

Units and Measurements · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 65 questions.

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Question 12 Marks
It is claimed that two cesium clocks, if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02s. What does this imply for the accuracy of the standard cesium clock in measuring a time-interval of 1s?
Answer
Error in 100 years = 0.02s Error in 1 sec $=\frac{0.02\text{s}}{100\times365\frac{1}{4}\times24\times60\times60}=\frac{2\times10^{-2}\times4}{1461\times24\times36\times10^4}$$7.9\times10^{-13}\text{s}\approx10^{-12}\text{s}.$
Hence, the accuracy of a standard caesium clock in measuring a time interval of 1s is $10^{-12}s$.
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Question 22 Marks
Answer the following: A screw gauge has a pitch of 1.0mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale ?
Answer
It is not possible to increase the accuracy of a screw gauge by increasing the number of divisions of the circular scale. Increasing the number divisions of the circular scale will increase its accuracy to a certain extent only.
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Question 32 Marks
Answer the following: The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only?
Answer
A set of 100 measurements is more reliable than a set of 5 measurements because random errors involved in the former are very less as compared to the latter.
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Question 42 Marks
Answer the following: You are given a thread and a metre scale. How will you estimate the diameter of the thread?
Answer
Wrap the thread on a uniform smooth rod in such a way that the coils thus formed are very close to each other. Measure the length of the thread using a meter scale. The diameter of the thread is given by the relation.
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Question 52 Marks
Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity): The number of strands of hair on your head.
Answer
Area of the head surface carrying hair = A With the help of a screw gauge, the diameter and hence, the radius of a hair can be determined. Let it be r.$\therefore$ Area of one hair $=\pi\text{r}^2$
Number of strands of hair $=\frac{\text{Total Surface area}}{\text{Area of one hair}}=\frac{\text{A}}{\pi\text{r}^2}$
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Question 62 Marks
Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity): The wind speed during a storm.
Answer
Wind speed during a storm can be measured by an anemometer. As wind blows, it rotates. The rotation made by the anemometer in one second gives the value of wind speed.
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Question 72 Marks
A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5mm. What is the estimate on the thickness of hair?
Answer
Magnification of the microscope = 100 Average width of the hair in the field of view of the microscope = 3.5mm$\therefore$ Actual thickness of the hair is 3.5/100 = 0.035mm
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Question 82 Marks
What is the technique used for measuring large time intervals?
Answer
For measuring large time intervals, we use the technique of radioactive dating. Large time intervals are measured by studying the ratio of number of radioactive atoms decayed to the number of surviving atoms in the specimen.
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Question 92 Marks
If x = at + bt, where x is in metre and t in hour, what will be the unit of 'a' and 'b'?
Answer
$x = at + bt^2$ So, the units of $\text{a}=\frac{\text{x}}{\text{t}}=\text{m}/\text{ hr}$ And $\text{b}=\frac{\text{x}}{\text{t}^2}=\text{m}/ (\text{hr})^2$
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Question 102 Marks
Write the order of following:
  1. Mass of a housefly.
  2. Mass of average man.
  3. Mass of an electron.
  4. Mass of earth.
Answer
i. $1 \times 10^{-4} \mathrm{~kg}$
ii. $7 \times 10^1 \mathrm{~kg}$
iii. $9.1 \times 10^{-31} \mathrm{~kg}$
iv. $6 \times 10^{24} \mathrm{~kg}$
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Question 112 Marks
The radius of a solid sphere is measured to be 11.24cm. What is the surface area of the sphere to appropriate significant figures?
Answer
r = 11.24cm Surface area $=4\pi\text{r}^2=4\pi(11.24) ^2$ $=4\times\frac{22}{7}\times(11.24)^2$ $=1588\text{cm}^2$
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Question 122 Marks
It is claimed that two cesium clocks, if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02s. What does this imply for the accuracy of the standard cesium clock in measuring a time-interval of 1s?
Answer
Error in 100 years = 0.02s Error in 1 sec $=\frac{0.02\text{s}}{100\times365\frac{1}{4}\times24\times60\times60}=\frac{2\times10^{-2}\times4}{1461\times24\times36\times10^4}$ $7.9\times10^{-13}\text{s}\approx10^{-12}\text{s}.$ Hence, the accuracy of a standard caesium clock in measuring a time interval of 1s is $10^{-12}s$.
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Question 132 Marks
The vernier scale of a travelling microscope has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5mm, calculate the minimum inaccuracy in the measurement of distance.
Answer
Here, parts on Vernier scale = n = 50 parts No. of division of M.S. coinciding with n parts of V.S. = (n - 1) $\therefore$ L.C. of instrument $=\frac{\text{L.C. of Mainscale}}{\text{No. of parts on V.S.}}=\frac{0.5\text{mm}}{50}$ Or minimum inaccuracy = 0.01mm.
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Question 142 Marks
Answer the following: A screw gauge has a pitch of 1.0mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale ?
Answer
It is not possible to increase the accuracy of a screw gauge by increasing the number of divisions of the circular scale. Increasing the number divisions of the circular scale will increase its accuracy to a certain extent only.
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Question 152 Marks
The viscous force 'F' acting on a body of radius 'r' moving with a velocity 'v' in a medium of coefficient of viscosity $'\eta'$ is given by $\text{F}=6\pi\eta\text{rv}$ Check the correctness of the formula.
Answer
$\text{F}=6\pi\eta\text{rv}$ $[\text{F}]=\text{MLT}^{-2}\dots(\text{i})$ $[\text{r}][\eta][\text{v}]=\text{L.ML}^{-1}\text{T}^{-1}\text{LT}^{-1}$ $=\text{MLT}^{-2}\dots{\text{(ii)}}$ since (i) and (ii) are equal. So the equation is correct.
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Question 162 Marks
The resistance R is the ratio of potential difference V and current I. What is the percentage error in R if V is $(100\pm5)\text{V}$ and I is $(10\pm2)$ A?
Answer
$\frac{\Delta\text{R}}{\text{R}}\times100=\pm\Big[\frac{\Delta\text{V}}{\text{V}}+\frac{\Delta\text{I}}{\Delta\text{I}}\Big]\times100$ $=\pm\Big[\frac{5}{100}+\frac{0.2}{10}\Big]\times100=\pm7\%$
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Question 172 Marks
Check the correctness of the relation $\text{v}^2-\text{u}^2=2\text{as}$ by method of dimensions. The symbol have their usual meaning.
Answer
The relation is given us $\text{v}^2-\text{u}^2=2\text{as}$ On LHS Dimension of $\text{v}^2=[\text{L}^2\text{T}^{-2}]$ And $\text{u}^2=[\text{LT}^{-1}]=[\text{L}^2\text{T}^{-2}]$ RHS $2\text{as}=[\text{LT}^{-2}][\text{L}]=[\text{L}^2\text{T}^{-2}]$ As dimensions of both terms on LHS are equal to the dimensions of RHS, the relation is dimensionally correct.
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Question 182 Marks
If $\text{A}=(12.0\pm0.1)\text{cm}$ and $\text{B}=(8.5\pm0.5)\text{cm},$ find:
  1. $\text{A}+\text{B}$
  2. $\text{A}-\text{B}$
Answer
  1. $\text{A}+\text{B}=20.5\pm0.6\text{cm}$
  2. $\text{A}-\text{B}=3.5\pm0.6\text{cm}$
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Question 192 Marks
Give the dimensional formula for surface energy, moment of inertia, angular velocity and gravitational force.
Answer
Dimensional formula for: Surface energy $=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$ Moment of inertia $=\left[\mathrm{ML}^2\right]$ Angular velocity $=\left[\mathrm{T}^{-1}\right]$ Gravitational force $=\left[\mathrm{MLT}^{-2}\right]$
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Question 202 Marks
If velocity of sound in a gas depends on its elasticity and density, derive the relation for the velocity of sound in a medium by the method of dimensions.
Answer
If u be the velocity of sound, E the elasticity of the medium and ρ the density of the medium, Then: $\text{v}\propto\text{E}^{\text{a}}\rho^{\text{b}}\text{ or }\upsilon\text{k}\text{ E}^{\text{a}}\rho^{\text{b}}\dots{\text{(i)}}$ Where k is a dimensionless constant of proportionality. Writing down the dimensions of both sides of equation (i), We get: $[\text{M}^0\text{LT}^{-1}]=[\text{ML}^{-1}\text{T}^{-2}]^{\text{a}}[\text{ML}^{-3}]^{\text{b}}$ $[\text{M}^0\text{LT}^{-1}]=[\text{M}^{\text{a}+\text{b}}\text{L}^{-a-3\text{b}}\text{T}^{-2\text{a}}]$ Comparing powers of M, L and T, We get $\text{a}+\text{b}=0$ $\text{-a}-3\text{b}=1$ $-\text{2a}=-1\text{ or }\text{a}=\frac{1}{2}$ $\therefore\frac{1}{2}+\text{b}=0\text{ or }\text{b}=\frac{-1}{2}$ From equations. (i) $\text{v}=\text{k}\text{ E}^{\frac{1}{2}}\rho^{\frac{-1}{2}}$ Or $\text{v}=\text{k}\sqrt{\frac{\text{E}}{\rho}}$ Where the value of k can be determined experimentally.
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Question 212 Marks
Is nuclear mass density dependent on the mass number? $(\text{Given}: \text{r} = \text{r}_0 \text{A}^{\frac{1}{3}})$
Answer
No, since density $=\frac{\text{Mass}}{\text{Volume}}=\frac{\text{A}}{\frac{4}{3}\pi\text{r}^3}=\frac{\text{A}}{\frac{4}{3}\pi\text{r}^3_0\text{A}}$ is independent of A.
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Question 222 Marks
The length and breadth of a rectangle are measured as $\text{}(\text{a}\pm\Delta\text{a})$ and $(\text{b}\pm\Delta\text{b})$ respectively. Find:
  1. Relative error.
  2. Absolute error in the measurement of area.
Answer
  1. Relative error in area:
$\frac{\Delta\text{A}}{\text{A}}=\Big[\frac{\Delta\text{a}}{\text{a}}+\frac{\Delta\text{b}}{\text{b}}\Big]$
$\text{as}\text{ A}=\text{ab}$
  1. Absolute error in area
$\Delta\text{A}=\Big(\frac{\Delta\text{a}}{\text{a}}+\frac{\Delta\text{b}}{\text{b}}\Big)\text{A}$
$=\Big(\frac{\Delta\text{a}}{\text{a}}+\frac{\Delta\text{b}}{\text{b}}\Big)\text{ab}$
$\Delta\text{A}=[(\Delta\text{a})\text{b}+(\Delta\text{b})\text{a}]$
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Question 232 Marks
If Length, Time and Energy are fundamental units, find the dimension of mass.
Answer
$\text{K.E.}=\text{E}=\frac{1}{2}\text{mv}^2$ $[\text{M}]=\Big[\frac{\text{E}}{\text{V}^2}\Big]=[\text{EL}^{-2}\text{T}^{+2}]$ $=[\text{ML}^2\text{T}^{-2}\text{L}^{-2}\text{T}^2]=[\text{M}]$
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Question 242 Marks
Using the principle of homogeneity of dimensions find which of the following is correct. where T is the time period, G is gravitational constant, M is mass and r is radius of orbit.
Answer
$[\text{T}^2] = \text{T}^2$$\Big[\frac{4\pi^2\text{r}^3}{\text{GM}}\Big] = \frac{\text{L}^3\text{M}^2}{\text{M}^2\text{LT}^2\text{L}^2}= \text{T}^2$
So, (iii) is correct.
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Question 252 Marks
Answer the following: The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only?
Answer
A set of 100 measurements is more reliable than a set of 5 measurements because random errors involved in the former are very less as compared to the latter.
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Question 262 Marks
Distinguish between dimensional variables and dimensional constants. Give example too.
Answer
Dimensional variables are those quantities which have dimensions and whose numerical value may change. Speed, velocity, acceleration etc., are dimensional variables. Dimensional constants are quantities having dimensions but having a constant value, e.g., gravitation constant (G), Planck's constant (H), Stefan's constant $(\sigma)$ etc.
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Question 272 Marks
The wavelength à associated with a moving particle depends upon its mass m, its velocity v and Planck's constant h. Show dimensionally the relationship between them.
Answer
$\lambda= \text{kh}^\text{a}\text{m}^\text{b}\text{v}^\text{c}$ $\Big[\text{M}^0\text{L}^1\text{T}^0\Big] = \Big[\text{ML}^2\text{T}^{-1}\Big]^\text{a}\Big[\text{M}^\text{b}\Big]\Big[\text{LT}^{-1}\Big]^\text{c}$ $=\text{M}^\text{a+b}\text{L}^{2\text{a+c}}\text{T}^\text{-a-c}$ Applying pruncipie of homogeneity of dimension: $\text{a}+\text{b}= 0$ $\text{2a}+\text{c}= 1$ $-\text{a}-\text{c}=0$ Solving the three equations, we get $\text{a}=1, $ $\text{b}=-1,$ $\text{c}=-1.$ $\lambda = \text{h}^1\text{m}^{-1}\text{v}^{-1}$ $\lambda = \frac{\text{h}}{\text{mv}}$
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Question 282 Marks
Find the height of a rock mountain, if the angle of elevation of its top increases from 30° to 45° on moving 100m towards the rock in the horizontal direction through the base of the rock.
Answer
$\text{In }\Delta\text{ABC},$ $\frac{\text{h}}{\text{x}}=\tan45^{\circ}\Rightarrow\text{h}=\text{x}$ $\text{In }\Delta\text{ABC},$ $\frac{​​\text{h}}{100+\text{x}}=\tan30^{\circ}$ $\frac{\text{h}}{100+\text{h}}=\frac{1}{\sqrt{3}}$ $\sqrt{3}\text{h}=100+\text{h}$ $\text{h}=\frac{100}{\sqrt{3}-1}$ $=50(\sqrt{3}+1)=136.5\text{m}$
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Question 292 Marks
If the length and time period of an oscillating pendulum have errors of 1% and 2% respectively, what is the error in the estimate of g?
Answer
We Know: $\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$ $\therefore\text{g}=4\pi2\frac{\text{l}}{\text{T}^2}$ $\frac{\Delta\text{g}}{\text{g}}=\frac{\Delta\text{l}}{\text{l}}+2\frac{\Delta\text{T}}{\text{T}}$ $\% \text{ error in g }=1\times+2\times2=5\%$
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Question 302 Marks
The length and breadth of a rectangle are measured as $(\text{a}\pm\Delta\text{a})$ and $(\text{b}\pm\Delta\text{b})$ respectively. Find:
  1. Relative error.
  2. Absolute error in the measurement of area.
Answer
Length $(\text{l})​​=\text{a}\pm\Delta\text{a}$
Breadth $(\text{b})=(\text{b}\pm\Delta\text{b})$
Area $\text{(A) = ab}$
Relative error $\pm\frac{\Delta\text{A}}{\text{A}}=\pm\frac{\Delta\text{a}}{\text{a}}\pm\frac{\Delta\text{b}}{\text{b}}$
Absolute error $\pm\Delta\text{A}=\pm\Big|\frac{\Delta\text{a}}{\text{a}}+\frac{\Delta\text{b}}{\text{b}}\Big|\text{ab}$
$\Rightarrow\pm\Delta\text{A}=[(\Delta\text{a})\text{b}+(\Delta\text{b})\text{a}]$
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Question 312 Marks
The length and breadth of a rectangle are $(5.7\pm0.1)\text{cm}$ and $(3.4\pm0.2)\text{cm}$ Calculate area of the rectangle with error limits.
Answer
$\text{l}=5.7\pm0.1\text{cm}$ $\text{b}=3.4\pm0.2\text{cm}$ $\text{Area}=\text{lb}=5.7\times3.4=19.38\text{cm}^2=19.4\text{cm}^2$ $\Delta\text{A}=\Big(\frac{\Delta\text{l}}{\text{l}}+\frac{\Delta\text{b}}{\text{b}}\Big)\times\text{A}$ $=\Big(\frac{0.1}{5.7}+\frac{0.2}{3.4}\Big )\times19.4$ $=(0.017+0.059)\times19.4$ $=1.47\simeq1.5 $ So, $\text{A}=(19.4\pm1.5)\text{cm}^2$
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Question 322 Marks
Write the order of following intervals in seconds:
  1. Time between two heart beats.
  2. Time of earth's revolution.
  3. Time of earth's rotation.
  4. Human life.
Answer
  1. $1\times 10^0\text{s}$
  2. $3\times 10^7\text{s}$
  3. $8.6\times10^4\text{s}$
  4. $2\times10^9\text{s}$
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Question 342 Marks
Why parallax method cannot be used for measuring distances of stars more than 100 light years away?
Answer
When a star is more than 100 light years away, then the parallax angle is so small that it cannot be measured accurately.
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Question 352 Marks
How many metric tons are there in teragram?
Answer
In 1 teragram $=10^2 \mathrm{~g} \ln 1$ metric ton $=10^3 \mathrm{~kg}=10^3 \times 10^3=10^6 \mathrm{~g} \therefore$ Number of metric tons are in teragram. $=\frac{10^{12}\text{g}}{10^{6}\text{g}}=10^6$
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Question 362 Marks
What is common between bar and torr?
Answer
Both bar and torr are the units of pressure. $1 \mathrm{~bar}=1$ atmospheric pressure $=760 \mathrm{~mm}$ of Hg column $=10^5 \mathrm{~N} / \mathrm{m}^2 1$ torr $=1 \mathrm{~mm}$ of Hg column $\therefore 1$ bar $=760$ torr
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Question 372 Marks
If the velocity of light is taken as the unit of velocity and year as the unit of time, what must be the unit of length? What is it called?
Answer
Unit of length = Unit of velocity $\times$ Unit of time $=3 \times 10^8 \mathrm{~ms}^{-1} \times 1$ year $=3 \times 108 \mathrm{~ms}^{-1} 365 \times 24 \times 60 \times 60 \mathrm{~s}=9.45 \times$ $10^{15} \mathrm{~ms}^{-1}=1$ light year
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Question 382 Marks
Moon is seen to be of $\Big(\frac{1}{2}\Big)^\circ$ diameter from the earth. What must be the relative size compared to the earth?
Answer
According to the problem, moon is seen as $\Big(\frac{1}{2}\Big)^\circ$ diameter from earth and earth is seen as $2^\circ$ diameter from moon. As $\theta$ is proportional to diameter, Hence, $\frac{\text{Diameter of earth}}{\text{Diameter of moon}}=\frac{2}{\Big(\frac{1}{2}\Big)}=4$
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Question 392 Marks
Find the value of one light year in giga metre.
Answer
We know that: $1\text{ ly}=9.46\times10^{15}\text{m}$ $1​​\text{Gm}=10^9\text{m}$ $\therefore1\text{ ly}=\frac{9.46\times10^{15}}{10^{9}}$ $=9.46\times10^{6}\text{Gm}$
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Question 402 Marks
Answer the following: You are given a thread and a metre scale. How will you estimate the diameter of the thread?
Answer
Wrap the thread on a uniform smooth rod in such a way that the coils thus formed are very close to each other. Measure the length of the thread using a meter scale. The diameter of the thread is given by the relation.
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Question 412 Marks
Check whether equation $\text{F.S}=\frac{1}{2}\text{mv}^2-\frac{1}{2}\text{mu}^2$ is dimensionally correct, where m is mass of the body, v its final velocity, u its initial velocity, F is force applied and S is the distance moved.
Answer
$\text{F.S}=\frac{1}{2}\text{mv}^2-\frac{1}{2}\text{mu}^2$ $\text{L.H.S}=[\text{ML}^2\text{T}^{-2}]$ $\text{R.H.S}=\frac{1}{2}[\text{M}][\text{LT}^{-1}]^{2}-\frac{1}{2}[\text{M}][\text{LT}^{-1}]$ $[\text{ML}^2\text{T}^{-2}]=[\text{ML}\text{T}^{-2}]-[\text{ML}^2\text{T}^{-2}]$ $\text{L.H.S}=\text{R.H.S}$ Which is dimensionally correct.
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Question 422 Marks
If $x = a + bt + ct^2$, where x is in metre and t in second, what are the units of a, b and c?
Answer
As $x = a + bt + ct^2$, where x is in metre and t in second. Hence, in accordance with the principle of homogeneity of dimensions, We have: Unit of a = x = metre Unit of b = unit of $\frac{\text{x}}{\text{t}}=\text{m}/\text{s}$ Unit of c = $\frac{\text{x}}{\text{t}^2}=\text{m}/(\text{s})^2$
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Question 432 Marks
Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity): The number of strands of hair on your head.
Answer
Area of the head surface carrying hair = A With the help of a screw gauge, the diameter and hence, the radius of a hair can be determined. Let it be r. $\therefore$ Area of one hair $=\pi\text{r}^2$ Number of strands of hair $=\frac{\text{Total Surface area}}{\text{Area of one hair}}=\frac{\text{A}}{\pi\text{r}^2}$
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Question 442 Marks
Rule out or accept the following formulae for kinetic energy on the basis of dimensional arguments.
  1. $\text{K}=\frac{3}{16}\text{mv}^2$
  2. $\text{K}=\frac{1}{2}\text{mv}^2+\text{ma}$
  3. $\text{K}=\frac{1}{2}\text{mv}^2$
Answer
Dimension of $\text{L.H.S.}$ and $\text{R.H.S.}$ are same in $(i)$ and $(iii).$ So they are accepted.
  1. $\text{K}=\frac{1}{2}\text{mv}^2+\text{ma}$ and $\Big[\frac{1}{2}\text{mv}^2\Big]\neq[\text{ma}].$
So it is wrong.
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Question 452 Marks
Magnitude of force F experienced by a certain object moving with speed v is given by $F = Kv^2$ where K is a constant. Find the dimensions of K.
Answer
$[\mathrm{F}]=\mathrm{MLT}^{-2}[\mathrm{v}]=\mathrm{LT}^{-1} \mathrm{~F}=\mathrm{Kv}^2\left[\mathrm{MLT}^{-2}\right]=\mathrm{K}\left[\mathrm{LT}^{-1}\right]^2$ $[\text{K}]=\frac{[\text{MLT}^{-2}]}{[\text{L}^2\text{T}^{-2}]}=[\text{ML}^{-1}]$
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Question 462 Marks
If displacement of a body is $\text{S}=(200\pm0.5)\text{m}$ and time taken by it is $\text{t}=(20\pm0.2)\text{s},$ then find the percentage error in the calculation of velocity.
Answer
Here, $\text{S}=200\text{m}$ $\Delta\text{S}=10.5\text{m},$ $\text{t}=20\text{s},$ $\Delta\text{t}=10.2\text{s}$ As, velocity $\text{v}=\frac{\text{s}}{\text{t}}$ $\therefore$ Percentage error in velocity $\frac{\Delta\text{v}}{\text{v}}=\Big[\frac{\Delta\text{S}}{\text{S}}+\frac{\Delta\text{t}}{\text{t}}\Big]\times100\%$ $=\Big[\frac{0.5}{200}+\frac{0.2}{20}\Big]\times100\%=1.25\%$
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Question 472 Marks
Calculate the solid angle subtended by the periphery of an area of $1\text{cm}^2$ at a point situated symmetrically at a distance of 5cm from the area.
Answer
Solid angle, $\Omega=\frac{\text{Area}}{(\text{Radial distance})^2}$ $=\frac{1\text{cm}^2}{(5\text{cm})^2}=\frac{1}{25}=4\times10^{-2}\ \text{steradian}$ ($\because$ Area = $1\text{cm}^2$, distance = 5cm)
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Question 482 Marks
The speed of sound in a solid is given by the formula: $\text{v}=\sqrt{\frac{\text{E}}{\rho}}$ Where, E is coefficient of elasticity and pis density of given solid. Check the relation by method of dimensional analysis.
Answer
In the given relation dimensions of LHS terms v are $\left[\mathrm{LT}^{-1}\right]$ Dimensional formula for $E$ and $p$ are $\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right)$ and $\left[\mathrm{ML}^{-3}\right]$ Dimesnsions of RHS $=\sqrt{\frac{\mathrm{ML}^{-1} \mathrm{~T}^{-1}}{\mathrm{ML}^{-3}}}=\sqrt{\mathrm{L}^2 \mathrm{~T}^{-2}}=\left[\mathrm{LT}^{-1}\right]$ As dimensions of LHS and RHS of the equation are same Hence the equation is dimensionally correct.
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Question 492 Marks
Write down the number of significant figures in the following:
  1. $5238N$
  2. $4200\ kg$
  3. $34.000m$
  4. $0.02340N/ m$
Answer
i. $5238 N$ has four significant digits.
ii. $4200 \mathrm{~kg}=4.200 \times 10^3 \mathrm{~kg}$ has four significant figures.
iii. $34.000 m$ has five significant digits.
iv. $0.02340 \mathrm{~N} / \mathrm{m}$ has four significant digits.
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Question 502 Marks
The mass of a proton is $1.67 \times 10^{-27} \mathrm{~kg}$. How many protons would make 1 g ?
Answer
Number of protons $=\frac{\text{Total mass}}{\text{mass of each proton}}$ $=\frac{10^{-3}}{1.67\times10^{-27}}$ $=5.99\times10^{23}$
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2 Marks Questions - Physics STD 11 Science Questions - Vidyadip