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?
AnswerError 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$.
View full question & answer→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 ?
AnswerIt 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.
View full question & answer→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?
AnswerA 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.
View full question & answer→Question 42 Marks
Answer the following: You are given a thread and a metre scale. How will you estimate the diameter of the thread?
AnswerWrap 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.
View full question & answer→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.
AnswerArea 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}$
View full question & answer→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.
AnswerWind 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.
View full question & answer→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?
AnswerMagnification 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
View full question & answer→Question 82 Marks
What is the technique used for measuring large time intervals?
AnswerFor 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.
View full question & answer→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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Write the order of following:
- Mass of a housefly.
- Mass of average man.
- Mass of an electron.
- Mass of earth.
Answeri. $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}$
View full question & answer→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?
Answerr = 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$
View full question & answer→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?
AnswerError 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$.
View full question & answer→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.
AnswerHere, 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.
View full question & answer→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 ?
AnswerIt 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.
View full question & answer→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.
View full question & answer→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\%$
View full question & answer→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.
AnswerThe 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.
View full question & answer→Question 182 Marks
If $\text{A}=(12.0\pm0.1)\text{cm}$ and $\text{B}=(8.5\pm0.5)\text{cm},$ find:
- $\text{A}+\text{B}$
- $\text{A}-\text{B}$
Answer
- $\text{A}+\text{B}=20.5\pm0.6\text{cm}$
- $\text{A}-\text{B}=3.5\pm0.6\text{cm}$
View full question & answer→Question 192 Marks
Give the dimensional formula for surface energy, moment of inertia, angular velocity and gravitational force.
AnswerDimensional 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]$
View full question & answer→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.
AnswerIf 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.
View full question & answer→Question 212 Marks
Is nuclear mass density dependent on the mass number? $(\text{Given}: \text{r} = \text{r}_0 \text{A}^{\frac{1}{3}})$
AnswerNo, 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.
View full question & answer→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:
- Relative error.
- Absolute error in the measurement of area.
Answer
- 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}$
- 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}]$ View full question & answer→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}]$
View full question & answer→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.
View full question & answer→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?
AnswerA 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.
View full question & answer→Question 262 Marks
Distinguish between dimensional variables and dimensional constants. Give example too.
AnswerDimensional 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.
View full question & answer→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, $ $\text{b}=-1,$ $\text{c}=-1.$ $\lambda = \text{h}^1\text{m}^{-1}\text{v}^{-1}$ $\lambda = \frac{\text{h}}{\text{mv}}$
View full question & answer→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$\text{In }\Delta\text{ABC},$ $\frac{\text{h}}{\text{x}}=\tan45^{\circ}\Rightarrow\text{h}=\text{x}$ $\text{In }\Delta\text{ABC},$ $\frac{\text{h}}{100+\text{x}}=\tan30^{\circ}$ $\frac{\text{h}}{100+\text{h}}=\frac{1}{\sqrt{3}}$ $\sqrt{3}\text{h}=100+\text{h}$ $\text{h}=\frac{100}{\sqrt{3}-1}$ $=50(\sqrt{3}+1)=136.5\text{m}$
View full question & answer→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?
AnswerWe Know: $\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$ $\therefore\text{g}=4\pi2\frac{\text{l}}{\text{T}^2}$ $\frac{\Delta\text{g}}{\text{g}}=\frac{\Delta\text{l}}{\text{l}}+2\frac{\Delta\text{T}}{\text{T}}$ $\% \text{ error in g }=1\times+2\times2=5\%$
View full question & answer→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:
- Relative error.
- Absolute error in the measurement of area.
AnswerLength $(\text{l})=\text{a}\pm\Delta\text{a}$
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 $\pm\Delta\text{A}=\pm\Big|\frac{\Delta\text{a}}{\text{a}}+\frac{\Delta\text{b}}{\text{b}}\Big|\text{ab}$
$\Rightarrow\pm\Delta\text{A}=[(\Delta\text{a})\text{b}+(\Delta\text{b})\text{a}]$
View full question & answer→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, $\text{A}=(19.4\pm1.5)\text{cm}^2$
View full question & answer→Question 322 Marks
Write the order of following intervals in seconds:
- Time between two heart beats.
- Time of earth's revolution.
- Time of earth's rotation.
- Human life.
Answer
- $1\times 10^0\text{s}$
- $3\times 10^7\text{s}$
- $8.6\times10^4\text{s}$
- $2\times10^9\text{s}$
View full question & answer→Question 332 Marks
If force F, length L and time T are taken as fundamental units then what be the dimensions of mass?
Question 342 Marks
Why parallax method cannot be used for measuring distances of stars more than 100 light years away?
AnswerWhen a star is more than 100 light years away, then the parallax angle is so small that it cannot be measured accurately.
View full question & answer→Question 352 Marks
How many metric tons are there in teragram?
AnswerIn 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. $=\frac{10^{12}\text{g}}{10^{6}\text{g}}=10^6$
View full question & answer→Question 362 Marks
What is common between bar and torr?
AnswerBoth 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
View full question & answer→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?
AnswerUnit 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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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?
AnswerAccording 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, $\frac{\text{Diameter of earth}}{\text{Diameter of moon}}=\frac{2}{\Big(\frac{1}{2}\Big)}=4$
View full question & answer→Question 392 Marks
Find the value of one light year in giga metre.
AnswerWe know that: $1\text{ ly}=9.46\times10^{15}\text{m}$ $1\text{Gm}=10^9\text{m}$ $\therefore1\text{ ly}=\frac{9.46\times10^{15}}{10^{9}}$ $=9.46\times10^{6}\text{Gm}$
View full question & answer→Question 402 Marks
Answer the following: You are given a thread and a metre scale. How will you estimate the diameter of the thread?
AnswerWrap 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.
View full question & answer→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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If $x = a + bt + ct^2$, where x is in metre and t in second, what are the units of a, b and c?
AnswerAs $x = a + bt + ct^2$, 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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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.
AnswerArea 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 $=\frac{\text{Total Surface area}}{\text{Area of one hair}}=\frac{\text{A}}{\pi\text{r}^2}$
View full question & answer→Question 442 Marks
Rule out or accept the following formulae for kinetic energy on the basis of dimensional arguments.
- $\text{K}=\frac{3}{16}\text{mv}^2$
- $\text{K}=\frac{1}{2}\text{mv}^2+\text{ma}$
- $\text{K}=\frac{1}{2}\text{mv}^2$
AnswerDimension of $\text{L.H.S.}$ and $\text{R.H.S.}$ are same in $(i)$ and $(iii).$ So they are accepted.
- $\text{K}=\frac{1}{2}\text{mv}^2+\text{ma}$ and $\Big[\frac{1}{2}\text{mv}^2\Big]\neq[\text{ma}].$
So it is wrong. View full question & answer→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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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.
AnswerHere, $\text{S}=200\text{m}$ $\Delta\text{S}=10.5\text{m},$ $\text{t}=20\text{s},$ $\Delta\text{t}=10.2\text{s}$ 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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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.
AnswerSolid 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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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.
AnswerIn 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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Write down the number of significant figures in the following:
- $5238N$
- $4200\ kg$
- $34.000m$
- $0.02340N/ m$
Answeri. $5238 N$ 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.
View full question & answer→Question 502 Marks
The mass of a proton is $1.67 \times 10^{-27} \mathrm{~kg}$. How many protons would make 1 g ?
AnswerNumber 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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Percentage error in the measurement of height and radius of cylinder are x and y respectively. Find percentage error in the measurement of volume. Which of the two measurements height or radius need more attention?
AnswerHeight of cylinder $=\text{x} $ Radius of cylinder$= \text{y}$ Volume of cylinder $\text{v}= \pi \text{y}^2\text{x}$ Percentage error in measurement of volime $\frac{\Delta\text{V}}{\text{V}}\times100 = \pm\Big(2\frac{\Delta\text{y}}{\text{y}}+\frac{\Delta\text{x}}{\text{x}}\Big)\times100$
Hence, radius needs more attention because any error in its measurement is multiplied two times.
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Rule out or accept the following formulae for kinetic energy on the basis of dimensional arguments:
- $\frac{3}{16}\text{mv}^2$
- $\frac{1}{2}\text{mv}^2+\text{ma}$
Answer$\text{K.E}=\frac{1}{2}\text{mv}^2$ Dimension of $\text{K}.\text{E}=[\text{M}\text{L}^2\text{T}^{-2}]$
Since Dimension of $K.E. =$ Dimension of $\frac{3}{16}\text{mv}^2$
It is dimensionally corrct. $\frac{1}{2}\text{mv}^2+\text{ma}$
Dimensions $=[\text{ML}^2\text{T}^{-2}]+[\text{MLT}^{-2}]$
Which is dimensionally incorrect.
View full question & answer→Question 532 Marks
A physical quantity X is given by: $\text{X}=\frac{\text{P}^2\text{Q}^{\frac{3}{2}}}{\text{R}^4\text{S}^{\frac{1}{2}}}$ The percentage errors in P, Q, R and S are 1%, 2%, 4% and 2%. Calculate the percentage error in X.
AnswerThe maximum fractional error in X is given by:$\frac{\Delta\text{X}}{\text{X}}=\pm\Big[\frac{2\Delta\text{P}}{\text{P}}+\frac{3}{2}\frac{\Delta\text{Q}}{\text{Q}}+\frac{4\Delta\text{R}}{\text{R}}+\frac{1}{2}\frac{\Delta\text{S}}{\text{S}}\Big]$
$\frac{\Delta\text{X}}{\text{X}}=\pm\Bigg[2\times\Big(\frac{1}{100}\Big)+\frac{3}{2}\Big(\frac{2}{100}\Big)+4\times\Big(\frac{4}{100}\Big)+\frac{1}{2}\times\Big(\frac{2}{100}\Big)\Bigg]$
$\frac{\Delta\text{X}}{\text{X}}=\pm\frac{2}{100}+\frac{3}{100}+\frac{16}{100}+\frac{1}{100}$
$=\pm\frac{22}{100}=\pm0.22$Percentage error in $\text{X}=\frac{\Delta\text{X}}{\text{X}}\times100$
$=\pm0.22\times100=\pm22%$
View full question & answer→Question 542 Marks
Name the device used for measuring the mass of atoms and molecules.
AnswerDeflection of a charge particle or ionized atom or molecule depends on the magnitude of either magnetic or electric field. Mass and Charge of a particle by using this principle can be measured by using spectrograph or spectrometer which measures the mass of atoms and molecules.
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For the determination of ‘g’ using a simple pendulum, measurements of I and T are required. Error in the measurement of which of these will have larger effect on the value of 'g' thus obtained and why? What is done to minimize this error?
AnswerError in measurement of time period T has larger effect on the value of g. Since $\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}}\Rightarrow\text{g}=\frac{\text{l}}{\text{T}^2}$ Thus, time for large number ofoscillations is measured to minimise error.
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The time of oscillation (t) of a small drop of liquid under surface tension depends upon the density $\rho,$ radius r and surface tension $\sigma.$ Prove dimensionally that to: $\text{t}\propto\sqrt{\frac{\rho\text{r}^3}{\sigma}}.$
AnswerTime of oscillation,$\text{t}\propto\rho^{\text{a}}\text{r}^{\text{b}}\sigma^{\text{c}}$
$\text{t}=\text{k}\rho^{\text{a}}\text{r}^\text{b}\sigma^\text{c},$ Where k = constant
Writing dimensions of both the sides, we have,$[\text{T}]=[\text{ML}^{-3}]^{\text{a}}[\text{L}]^{\text{b}}[\text{ML}^2]^\text{c}$
$[\text{M}^{\text{a}+\text{c}}\text{L}^{-3\text{a}+\text{b}}\text{T}^{-2\text{c}}]$
Comparing the powers of M, L and T on both sides, We have,$\text{a}+\text{c}=0\dots(\text{i})$
$-3\text{a}+\text{b}=0\dots(\text{ii})$
$-2\text{c}=1\dots(\text{iii})$
$\text{a}=\frac{1}{2},\text{c}=-\frac{1}{2}\text{ and}\text{ b}=\frac{3}{2}$
Putting these values in,$\text{t}=\text{k}\rho^\text{a}\text{r}^\text{b}\sigma^\text{c}$
We get $\text{t}=\text{k}\rho^{\frac{1}{2}}\text{r}^{\frac{3}{2}}\sigma^{\frac{-1}{2}}$
OR $\text{t}\propto\sqrt{\frac{\rho\text{r}^3}{\sigma}}.$
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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.
AnswerWind 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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The mean value of period of oscillation of a simple pendulum in an experiment is 2.62s. The arithmetic mean of all the absolute errors is 0.11s. Round off the period of simple pendulum to appropriate number of significant figures with reasons.
Answer$\text{T}=2.62\text{ sec}.\overline{\text{T}}=0.11\text{ sec}$ $\Rightarrow\text{T}=2.62\pm0.11\text{ sec}$ Error is 0.11, which is $\frac{1}{10}\text{th}$ Hence T = 2.6 seconds uncertain digit is the '6'. So, with significance figures we write T = 2.6 seconds.
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Which of the following length measurement is most accurate and why?
- $4.00\ cm.$
- $0.004\ mm.$
- $40.00\ cm.$
Answer
- $\frac{\Delta\text{x}}{\text{x}}=\frac{0.01}{4.00}=0.0025$
- $\frac{\Delta\text{x}}{\text{x}}=\frac{0.001}{0.004}=0.25$
- $\frac{\Delta\text{x}}{\text{x}}=\frac{0.01}{40.00}=0.00025$
The last observation has the least fractional error and hence, it is more accurate. View full question & answer→Question 602 Marks
Out of formulae:
- $\text{y}=\text{a}\sin\frac{2\pi\text{t}}{\text{T}}$
- $\text{y}=\text{a}\sin\text{vt}$
for the displacement $y$ of a particle undergoing a certain periodic motion, rule out the wrong formula on dimensional grounds. $[$where $a =$ maximum displacement of the particle, $v =$ speed of the particle, $T =$ time period of motion$.]$ Answer$\frac{\text{2}\pi\text{t}}{\text{T}}$ have dimensions $\text{M}^{0}\text{L}^{0}\text{T}^{0}$
$\therefore\text{y}=\text{a}$ Hence correct.
$vt$ have dimensions $\text{M}^{0}\text{L}^{1}\text{T}^{0}$ not dimensionless, so it is wrong.
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The radius of atom is of the order of $2\mathring{\text{A}}$ and radius of a nucleus is of the order of fermi. How many magnitudes higher is the volume of atom as compared to the volume of nucleus?
Answer$R_A$, i.e. radius of atom is $2\mathring{\text{A}}=2\times10^{-10}\text{m}$ $R_N$, i.e. radius of nucleas is 1 fermi $=10^{-15}\text{m}$ $\frac{\text{V}_\text{A}}{\text{V}_\text{N}}=\frac{\frac{4}{3}\pi\text{R}^3\text{A}}{\frac{4}{3}\pi\text{R}^3\text{N}}=\Big[\frac{\text{R}_{\text{A}}}{\text{R}_\text{N}}\Big]^3$ $=\Big[\frac{2\times10^{-10}}{10^{-15}}\Big]=8\times10^{15}$
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The distance of a galaxy is of the order of $10^{25}m$. Calculate the order of magnitude of time taken by light to reach us from the galaxy.
AnswerAccording to the problem, distance of the galaxy = $10^{25}m$. Speed of light = $3 \times 10^8m/s$ Hence, time taken by light to reach us from galaxy is, $\text{t}=\frac{\text{Distance}}{\text{Speed}}=\frac{10^{25}\text{m}}{3\times10^8\text{m/s}}=\frac{1}{3}\times10^{17}$ $=\frac{10}{3}\times10^{16}=3.33\times10^{16}\text{s}$
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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?
AnswerMagnification 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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Each side of a cube is 7.203m. Calculate the surface area and volume of the cube upto correct number of significant figure.
AnswerSignificant figure in length $=4 \therefore$ Significant figure in surface area and volume $=4$ Surface area of cube $=$ $6(7.203)^2 \mathrm{~m}^2=311.299254=311.3 \mathrm{~m}^2$ Volume of cube $=(7.203)^3 \mathrm{~m}^3=373.714754 \mathrm{~m}^2=373.7 \mathrm{~m}^3$
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The SI unit of energy is $J = kg m ^2 s ^{-2}$; that of speed $v$ is $m s ^{-1}$ and of acceleration $a$ is $m s ^{-2}$. Which of the formulae for kinetic energy ( $K$ ) given below can you rule out on the basis of dimensional arguments ( $m$ stands for the mass of the body) :
(a) $K=m^2 v^3$
(b) $K=(1 / 2) m v^2$
(c) $K=m a$
(d) $K=(3 / 16) mv ^2$
(e) $K=(1 / 2) m v^2+m a$
AnswerEvery correct formula or equation must have the same dimensions on both sides of the equation. Also, only quantities with the same physical dimensions can be added or subtracted. The dimensions of the quantity on the right side are $\left[ M ^2 L ^3 T ^{-3}\right]$ for (a); $\left[ M L ^2 T ^{-2}\right]$ for (b) and (d); $\left[ M L T ^{-2}\right]$ for (c). The quantity on the right side of (e) has no proper dimensions since two quantities of different dimensions have been added. Since the kinetic energy $K$ has the dimensions of [ $\left.ML ^2 T ^{-2}\right]$, formulas (a), (c) and (e) are ruled out. Note that dimensional arguments cannot tell which of the two, (b) or (d), is the correct formula. For this, one must turn to the actual definition of kinetic energy (see Chapter 5). The correct formula for kinetic energy is given by (b).
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