By QB365 on 31 Dec, 2022
QB365 provides a detailed and simple solution for every Possible Questions in Class 11 Physics Subject - Important 2 Mark English Medium. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.
11th Standard
Physics
Answer All The Questions
Write the rules for determining significant figures.
Explain what is meant by Cartesian coordinate system?
A particle has its position moved from \(\overset { \rightarrow }{ { r }_{ 1 } } =3\hat { i } +4\hat { j } \) to \(\overset { \rightarrow }{ { r }_{ 2 } } =\hat { i } +2\hat { j } \) Calculate the displacement vector (\(\Delta \)\(\overrightarrow { r } \)) and draw the \(\overrightarrow { { r }_{ 1 } } \), \(\overrightarrow { { r }_{ 2 } } \) and \(\Delta \overrightarrow { r } \) vector in a two dimensional cartesian coordinate system.
If the position vector of the particle is given by \(\vec { r } ={ 3t }^{ 2 }\hat { i } +5t\hat { j } +4\hat { k } \), Find the
(a) The velocity of the particle at t = 3 s
(b) Speed of the particle at t = 3 s
(c) Acceleration of the particle at time t = 3s
The position vector of a particle is given by \(\vec { r } =3t\hat { i } +5t^{ 2 }\hat { j } +7\hat { k } \). Find the direction in which the particle experiences net force?
Consider an object of mass 50 kg at rest on the floor. A Force of 5 N is applied on the object but it does not move. What is the frictional force that acts on the object?
Explain the characteristics of elastic and inelastic collision.
Define torque and mention its unit.
Four round objects namely a ring, a disc, a hollow sphere and a solid sphere with same radius R start to roll down an incline at the same time. Find out which object will reach the bottom first.
If the Earth has no tilt, what happens to the seasons of the Earth?
Within the elastic limit, the stretching strain produced in wires A, B, and C due to stress is shown in the figure. Assume the load applied are the same and discuss the elastic property of the material.
Write down the elastic modulus in ascending order
State Pascal’s law in fluids.
a. ‘A lake has more rain’.
b. ‘A hot cup of coff ee has more heat’.
What is wrong in these two statements?
Eiffel tower is made up of iron and its height is roughly 300 m. During winter season (January) in France the temperature is 2°C and in hot summer its average temperature 25°C. Calculate the change in height of Eiffel tower between summer and winter. The linear thermal expansion coefficient for iron α = 10 x 10-6 per °C.
500 g of water is heated from 30°C to 60°C. Ignoring the slight expansion of water, calculate the change in internal energy of the water? (specific heat of water 4184 J/kg.K)
Can we measure the temperature of the object by touching it?
In an adiabatic expansion of the air, the volume is increased by 4%, what is percentage change in pressure?
(For air \(\gamma \) = 1.4)
Suppose a person wants to increase the efficiency of the reversible heat engine that is operating between 100°C and 300°C. He had two ways to increase the efficiency. (a) By decreasing the cold reservoir temperature from 100°C to 50°C and keeping the hot reservoir temperature constant (b) by increasing the temperature of the hot reservoir from 300°C to 350°C by keeping the cold reservoir temperature constant. Which is the suitable method?
Which of the following functions of time represent periodic and non-periodic motion?
a. sin ωt + cos ωt
b. ln ωt
Show that for a particle executing simple harmonic motion
a. the average value of kinetic energy is equal to the average value of potential energy.
b. average potential energy = average kinetic energy = \(\frac{1}{2}\) (total energy)
Hint: average kinetic energy =< kinetic energy > = \(\frac{1}{T} \int_{0}^{T}\)(Kineticenergy) dt and
average Potential energy = < Potential energy > = \(\frac{1}{T} \int_{0}^{T}\) (Potential energy) dt
A force \(\vec { F } =(6\hat { i } -8\hat { j } +10\hat { k } )\) N produces acceleration of 1 m/s2 in a body. Calculate the mass of the body.
Wheels are made circular. Why?
Three blocks of masses 10 kg, 7 kg and 2 kg are placed in contact with each other on a frictionless table. A force of 50 N is applied on the heaviest mass. What is the acceleration of the system?
Which law of motion is involved in rocket propulsion?
A car starts from rest and moves on a surface with uniform acceleration. Draw the graph of kinetic energy versus displacement. What information you can get from that graph?
A light body and a heavy body have the same linear momentum. Which one has greater K.E?
Define power.
A labourer standing near the top of an old wooden step ladder feels unstable. Why?
If the K.E of a satellite revolving around the earth in any orbit is doubled then what will happen to satellite?
Whistle of approaching railway engine is shriller than the receding engine. Give reason
Answers
Rule | Example |
(i) All non-zero digits are significant | 1342 has four significant figures |
(ii) All zeros between two non zero digits are significant | 2008 has four significant figures |
(iii) All zeros to the right of a non-zero digit but to the left of a decimal point are significant. | 30700 has five significant figures |
(iv) a) The number without a decimal point, the terminal or trailing zero( s) are not significant. | 30700 has three significant figures |
b) All zeros are significant if they come from a measurement | 30700 has three significant figures |
(v) If the number is less than 1, the zero (s) on the right of the decimal point but to left of the first non zero digit are not significant. | 0.00345 has three significant figures |
(vi) All zeros to the right of a decimal point and to the right of non-zero digit are significant. | 40.00 has four significant figures and 0.030400 has five significant figures |
(vii) The number of significant figures does not depend on the system of units used | 1.53 cm, 0.0153 m, 0.0000153 km, all have three significant figures. |
(i) Cartesian Coordinate system is a frame of reference in which the position of an object at any given instant is described in terms of its distances along x, y and z axes.
(ii) Conventionally right - handed Cartesian Coordinate system where the x, y and z axes are drawn in anti clockurise direction is followed in physics.
\(\vec { { r }_{ 1 } } =3\hat { i } +4\hat { j } \)
\(\vec { { r }_{ 2 } } =\hat { i } +2\hat { j } \)
\(\Delta \overrightarrow { r } =\overrightarrow { { r }_{ 2 } } -\overrightarrow { { r }_{ 1 } } \)
\(=(\hat { i } +2\hat { j } )-(3\hat { i } +4\hat { j } )\)
\(\Delta \overrightarrow { r } =-2\hat { i } +\hat { j } \)
(a) The velocity
\(\vec { v } =\frac { d\vec { r } }{ dt } =\frac { dx }{ dt } \hat { i } +\frac { dy }{ dt } \hat { j } +\frac { dz }{ dt } \hat { k } \)
We obtain \(\vec { v } (t)=6t\hat { i } +5\hat { j } \)
The velocity has only two components vx = 6t, depending on time t and vy = 5 which is independent of time.
The velocity at t = 3 s is \(\vec { v } \)(3) =\(18\hat { i } +5\hat { j } \)
(b) The speed at t = 3 s is
\(v=\sqrt { { 18 }^{ 2 }+5^{ 2 } } =\sqrt { 349 } \approx \) 18.68 m s-1
(c) The acceleration \(\vec { a } \) is \(\vec { a } =\frac { { d }^{ 2 }\vec { r } }{ { dt }^{ 2 } } =6\hat { i } \)
The acceleration has only the x-component. Note that acceleration here is independent of t, which means \(\vec { a } \) is constant. Even at t = 3 s it has same value a = 6t. The velocity is non uniform, but the acceleration is uniform (constant) in this case.
Velocity of the particle,
\(\vec { v } =\frac { d\vec { r } }{ dt } =\frac { d }{ dt } (3t)\hat { i } +\frac { d }{ dt } (5{ t }^{ 2 })\hat { j } +\frac { d }{ dt } (7)\hat { k } \)
\(\frac { d\vec { r } }{ dt } =3\hat { i } +10t\hat { j } \)
Acceleration of the particle
\(\frac { d\vec { r } }{ dt } =\frac { { d }\vec { v } }{ dt } =\frac { { d }^{ 2 }\vec { r } }{ dt^{ 2 } } =10\hat { j } \)
Here, the particle has acceleration only along positive y direction. According to Newton's second law, net force must also act along positive y direction. In addition, the particle has constant velocity in positive x direction and no velocity in z direction. Hence, there are no net force along x or z direction.
When the object is at rest, the external force and the static frictional force are equal and opposite.
The magnitudes of these two forces are equal fs=Fext
Therefore, the static frictional force acting on the object is
fs = 5N
The direction of this frictional force is opposite to the direction of Fext.
Characteristics of elastic collision are
1. Total momentum remains conserved
2. Total kinetic energy remains conserved.
3. In elastic collision conservative forces are involved. Hence total kinetic energy is conserved.
4. In elastic collision, mechanical energy is not dissipated.
Characteristics of inelastic collision are
1. Total momentum is conserved.
2. Total kinetic energy is not conserved.
3. Forces involved are non-conservative forces
4. Mechanical energy is dissipated into heat, light, sound etc.
Torque is defined as the moment of the external applied force about a point or axis of rotation.
Torque \(\overrightarrow { \tau } =\overrightarrow { r } \times \overrightarrow { F } \) its unit is Nm.
For all the four objects namely the ring, disc, hollow sphere and solid sphere, the radii of gyration K are R,\(\sqrt{\frac{1}{2}R},\sqrt{\frac{2}{3}R},\sqrt{\frac{2}{5}R}\). With numerical values the radius of gyration K are 1R, 0.707R, 0.816R, 0.632R respectively. The expression for time taken for rolling has the radius of gyration K in the numerator as per equation.
\(t=\sqrt{\frac{2h(1+\frac{K^{2}}{R^{2}})}{8 sin ^{2}\theta}}\) \(V=\sqrt{\frac{2gh}{[1+\frac{K^{2}}{R^{2}}]}}\)
The one with least value of radius of gyration K will take the shortest time to reach the bottom of the inclined plane. The order of objects reaching the bottom is first, solid sphere second, disc third, hollow sphere and last, ring.
If the Earth has us tilt then there would not be seasons of the Earth.
Here, the elastic modulus is Young modulus and due to stretching, stress is tensile stress and strain is tensile strain.
Within the elastic limit, stress is proportional to strain (obey Hooke’s law). Therefore, it shows a straight line behaviour. So, Young modulus can be computed by taking slope of these straight lines. Hence, calculating the slope for the straight line, we get
Slope of A > Slope of B > Slope of C
Which implies,
Young modulus of C < Young modulus of B < Young modulus of A
Notice that larger the slope, lesser the strain (fractional change in length). So, the material is much stiffer. Hence, the elasticity of wire A is greater than wire B which is greater than C. From this example, we have understood that Young’s modulus measures the resistance of solid to a change in its length.
Pascal's law states that, if the pressure in a liquid is changed at a particular point, the change is transmitted to the entire liquid without being diminished in magnitude.
a. When it rains, lake receives water from the cloud. Once the rain stops, the lake will have more water than before raining. Here ‘raining’ is a process which brings water from the cloud. Rain is not a quantity rather it is water in transit. So the statement ‘lake has more rain is wrong, instead the ‘lake has more water will be appropriate.
b. When heated, a cup of coffee receives heat from the stove. Once the coffee is taken from the stove, the cup of coffee has more internal energy than before. ‘Heat’ is the energy in transit and which flows from an object at higher temperature to an object at lower temperature. Heat is not a quantity. So the statement ‘A hot cup of coffee has more heat is wrong, instead ‘coffee is hot’ will be appropriate.
\(\frac { \Delta L }{ L } ={ \alpha }_{ L }{ \Delta }_{ T }\)
\(\Delta L={ L\alpha }_{ L }{ \Delta }_{ T }\)
\(\Delta\)L = 10\(\times\)10-6 \(\times\)300\(\times\)23 = 0.69 m=69 mm
Area Expansion
For a small change in temperatur ΔT the fractional change in area \(\left(ΔA\over A_0\right)\) substance is directly proportional to ΔT and it can be written as
\({ΔA\over A_0}=\alpha_AΔT\)
Therefore, \(\alpha_A={ΔA\over A_0ΔT}\)
Where, αA = coefficient of area expansion.
ΔA = Change in area
A0 = Original area
ΔT = Change in temperature
Volume Expansion
For a small change in temperature ΔT the fractional change in volume \(\left(ΔV\over V_0\right)\) of a substance is directly proportional to ΔT.
\({ΔV\over V_0}=\alpha_VΔT\)
Therefore, \(\alpha_V={ΔV\over V_0ΔT}\)
Where, αV = coefficient of volume expansion.
ΔV = Change in volume
V0 = Original volume
ΔT = Change in temperature
Unit of coeffi cient of linear, area and volumetric expansion of solids is oC-1 or K-1
When the water is heated from 30°C to 60°C, there is only a slight change in its volume. So we can treat this process as isochoric. In an isochoric process the work done by the system is zero. The given heat supplied is used to increase only the internal energy.
ΔU = Q = msv ΔT
The mass of water = 500 g = 0.5 kg
The change in temperature = 30K
The heat Q = 0.5\(\times\)4184\(\times\)30 = 62.76 kJ
We can measure the temperature of the object by touching it. By the way of conduction the heat from a body is transferred to the mercury, in a thermometer. According to the heat of the body, mercury will rise in the thermometer that indicates the temperature.
When volume is increased by 4 %
\(\frac{\Delta V}{V} \times 100=4 \% \quad \gamma=1.4
\)
\(\frac{\Delta P}{P}+\gamma \frac{\Delta V}{V}=0
\)
\(\gamma \frac{\Delta V}{V} \times 100=\left|-\frac{\Delta P}{P}\right|=\frac{\Delta P}{P}
\)
\( \therefore \text { Percentage of decrease in pressure }=\frac{\Delta P}{P} \times 100
\)
\( \frac{\Delta P}{P} \times 100=\left(\frac{\Delta V}{V} \times 100\right) \times 4
\)
\( \frac{\Delta P}{P} \times 100=1.4 \times 4=5.6 %\)
∴ Percentage is decreased by = 5.6 %
Temperature of sink \(T_{L}=100+273=373 \mathrm{~K}\)
Temperature of source \(\mathrm{T}_{\mathrm{H}}=300+273=573 \mathrm{~K}\)
Efficiency \(\eta =1-\frac{373}{573}
\)
\(=\frac{573-373}{573}=\frac{200}{573}=0.349=34.9 \%\)
Method (a)
By decreasing the temperature \(\mathrm{T}_{\mathrm{L}} \ from \ 100^{\circ} \mathrm{C} \ to \ 50^{\circ} \mathrm{C}\)
\(\mathrm{T}_{\mathrm{L}} =50+273=323 \mathrm{~K}
\)
\(\mathrm{~T}_{\mathrm{H}} =300+273=573 \mathrm{~K}
\)
\(\therefore\) Efficiency \(\eta =1-\frac{323}{573}
\)
\(=\frac{573-323}{573}=\frac{250}{573}=0.4363=43.63 \%\)
\(\therefore\) Efficiency \(\eta=1-\frac{323}{573}\)
Method (b)
By increasing the temperature \(\mathrm{T}_{\mathrm{H}}\ from \ 300^{\circ} \mathrm{C} \ to \ 350^{\circ} \mathrm{C}.\)
\(\mathrm{T}_{\mathrm{L}} =100+273=373 \mathrm{~K}
\)
\(\mathrm{T}_{\mathrm{H}} =350+273=623 \mathrm{~K}
\)
\(\therefore\) Efficiency \(\eta =1-\frac{T_{L}}{T_{H}}
\)
\(=1-\frac{373}{623}=\frac{623-373}{623}=\frac{250}{623}=0.4012 \\
\)
\(=40.12 \%\)
\(\therefore \text {Efficiency } \quad \eta=1-\frac{T_{L}}{T_{H}}\)
Hence Method (a) is more important than method (b). Hence method (a) is suitable method.
a. Periodic
b. Non-periodic
(a) Suppose a particle of mass m executes SHM of time period T. The displacement of the particle of any instant A is given by
\(\mathrm{y} =\mathrm{A} \sin \omega t \) .....(1)
\(\text {Velocity } \mathrm{v} =\frac{d y}{d t}=\frac{d}{d t}\)
\((\mathrm{A} \sin \omega t) =\omega A \cos \omega t\)
\( \therefore \text {Kinetic energy } E_{k} =\frac{1}{2} m v^{2}=\frac{1}{2} m \omega^{2} \cdot A^{2} \cos ^{2} \omega t \) ......(2)
Potential energy \(\mathrm{E}_{\mathrm{p}}=\frac{1}{2} m \omega^{2} y^{2}\)
\(\mathrm{E}_{p}=\frac{1}{2} m \omega^{2} A^{2} \sin ^{2} \omega t\) .....(3)
Average K.E for one period of oscillation is
\(E_{k_{\alpha v}}=\frac{1}{T} \int_{0}^{T} E_{k} d t \)
\(=\frac{1}{T} \int \frac{1}{2} m \omega^{2} A^{2} \cos ^{2} \omega t d t \)
\(=\frac{1}{2 T} m \omega^{2} A^{2} \int_{0}^{T} \frac{(1+\cos 2 \omega t)}{2} d t \)
\(E_{k_{o v}}=\frac{1}{4 T} m \omega^{2} A^{2}\left[t+\frac{\sin 2 \omega t}{2 \omega}\right]_{0}^{T}=\frac{1}{4 T} m \omega^{2} A^{2}(T) \)
\(E_{k_{\alpha v}}=\frac{1}{4 T} m \omega^{2} A^{2}\) .....(4)
Average potential energy over a period of oscillation is
\(E_{p_{\omega}} =\frac{1}{T} \int_{0}^{T} E_{p} d t=\frac{1}{T} \int_{0}^{T} \frac{1}{2} m \omega^{2} A^{2} \sin ^{2} \omega t d t \)
\(=\frac{1}{2 T} m \omega^{2} A^{2} \int_{0}^{T}\left(\frac{1-\cos 2 \omega t}{2}\right) d t \)
\(=\frac{1}{4 T} m \omega^{2} A^{2}\left[t-\frac{\sin 2 \omega t}{2 \omega}\right]_{0}^{T}=\frac{1}{4 T} m \omega^{2} A^{2} C_{1} \)
\(E_{p_{a v}} =\frac{1}{4 T} m \omega^{2} A^{2} \) .....(5)
clearly from equation (4) and (5)
\(E_{k_{\mathrm{ow}}}=E_{p_{\mathrm{wo}}}\)
Average value of K.E = Average value of P.E
(b) Total energy
\(\text {T.E } =\frac{1}{2} m \omega^{2} y^{2}+\frac{1}{2} m \omega^{2}\left(A^{2}-y^{2}\right) \)
\(\text {But y } =\mathrm{A} \sin \omega t \)
\(\text {T.E } =\frac{1}{2} m \omega^{2} A^{2} \sin ^{2} \omega t+\frac{1}{2} m \omega^{2} A^{2} \cos ^{2} \omega t \)
\(=\frac{1}{2} m \omega^{2} A^{2}\left(\sin ^{2} \omega t+\cos ^{2} \omega t\right)\)
\(But \sin ^{2} \omega t+\cos ^{2} \omega t=1\)
Equation (6) becomes
\(\text {T.E }=\frac{1}{2} m \omega^{2} A^{2}(1)=\frac{1}{2} m \omega^{2} A^{2}\)
Average potential energy = Average kinetic energy
\(=\frac{1}{2} \text { (Total energy) }\)
Given, \(\vec { F } =(6\hat { i } -8\hat { j } +10\hat { k } )\)
m = \(\frac { |\vec { F } | }{ a } =\frac { \sqrt { { 6 }^{ 2 }+{ 8 }^{ 2 }+{ 10 }^{ 2 } } }{ 1 } =10\sqrt { 2 } \) kg
Rolling friction is less than sliding friction.
We know that a=\(\left[ \frac { F }{ { m }_{ 1 }+{ m }_{ 2 }+{ m }_{ 3 } } \right] =\frac { 50N }{ 10kg+7kg+2kg } =\frac { 50 }{ 19 } \)= 2.63 ms-2
Newton's third law of motion.
K.E=\(\frac{1}{2}\)mv2
=\(\frac{1}{2}\)m(2 as) [From equation of motion II]
∴ K.E.α S.
The lighter body has more K.E. as K.E. = \(\frac { { p }^{ 2 } }{ 2m } \) and for constant p, K.E.\(\propto \frac { 1 }{ m } \) .
Power is defined as the rate of work done or energy delivered.
Power(P)=\(\frac{work\ done(W)}{time\ taken(t)}\)
The ladder can rotate about the point of contact of the ladder with the ground. When the labourer is at the top of the ladder, the lener arm of the force is large. Hence the turning effect on the ladder will be large.
(i) The total energy of a satellite in any orbit E=-k.
(ii) If kinetic energy is doubled, i.e an additional kinetic energy is given to it.
∴ E = -k + k = 0 and the satellite will leave its orbit and go to infinity
This is due to Doppler effect. When the source of sound (engine) moves towards the stationary observer with velocity vs, then the apparent frequency is
\(f_{1}^{\prime}=\left(\frac{v}{v-v_{s}}\right) f\)
When the source recedes from a stationary observer with velocity v, then the apparent frequency is
\(f_{2}^{\prime}=\left(\frac{v}{v+v_{s}}\right) f \quad \text { where } v \text { - velocity of sound. }\)
It is clear that \(f_{1}^{\prime}>f_{2}^{\prime}\). Hence the whistle of the approaching railway engine is shorter than the receding engine