## CBSE - MCQ Question Banks (के. मा. शि. बो . -प्रश्नमाला )

Q. 201 1 light year is equal to

A. 9.123 Χ102m.

B. 8.235 Χ105m.

C. 9.467 Χ1015m.

D. 2.276 Χ1010m.

#### SOLUTION

1 light year is the distance traveled by light in vacuum in one year. 1 light year = 3 x 108 x (365 x 24 x 60 x 60) m = 9.46 x 1015 m

Q. 202 A quantity has dimensions. Is it necessary it must have a unit?

#### SOLUTION

Yes If a quantity has dimensions then it must have a unit.

Q. 203 What is principle of homogeneity in dimensional method?

#### SOLUTION

It states that the dimensions of each term on both sides of the equation should remain same.

Q. 204 Which method is used for measuring nuclear sizes?

#### SOLUTION

Rutherford method

Q. 205 What is atomic mass unit (a.m.u.)?

#### SOLUTION

It is defined as (1/12)th of the mass of one 6C12 atom.

Q. 206 Name the error associated with the resolution of the instrument.

#### SOLUTION

Least count Error

Q. 207 What is the responsibility of National physical laboratory (NPL)?

#### SOLUTION

The National physical Laboratory has the responsibility of maintenance and improvement of physical standards, including that of time, frequency etc.

Q. 208 What is the significance of precision?

#### SOLUTION

Precision tells us to what resolution or limit the quantity is measured

Q. 209 The  parallax of a heavenly body measured from two points diametrically opposite on the equator of the earth is 1.0 minute. If the radius of the earth is 6400 km, find the distance of the heavenly body from the centre of the earth in AU. Take 1 AU = 1.5 × 1011 m.

A.  0.923

B. 0.293

C. 0.329

D. 0.429

#### SOLUTION

We know that
S = D/ where, D is the diameter of the earth Here,  D = 2  6400 km;

Q. 210 The random errors can be reduced by

A. Taking more number of observation

B. Eliminating the error

C. By taking more care

D. By ignoring the least count (if any) of the instrument

#### SOLUTION

Random errors cannot be eliminated altogether even after taking atmost care. It can only be reduced by taking more number of observations

Q. 211 Which of the following has meter kelvin as the unit ?

A. Rydberg constant

B. Wein's constant

C. Solar constant

D. Gas constant

#### SOLUTION

As Wein's constant, b = T, therefore its unit is meter kelvin.

Q. 212 The dimension equation is [M0L0T-1]. The physical quantity associated is

A. frequency

B. velocity

C. time

D. Heat

#### SOLUTION

Frequency = 1 / time

Q. 213 A proton has a charge of

A. +1.6 x 10-19 Coulomb

B. -1.6 x1018 Coulomb

C. +1.6 x 10+18 Coulomb

D. -2.6 x 10+19 Coulomb

#### SOLUTION

Proton charge is 1.6 x 10-19 Coulomb and Electron charge is -1.6 x 10-19 Coulomb

Q. 214 Both 'Light year' and 'Year' are measured in

A. meter and second respectively.

B. only second.

C. only meter.

D. only minute.

#### SOLUTION

Since, Light year is a length and Year is time.

Q. 215 The dimension equation is [M0L0T-1]. The physical quantity associated is

A. frequency.

B. velocity.

C. time.

D. heat.

#### SOLUTION

Frequency = 1 / time

Q. 216 The random errors can be reduced by

A. taking more number of observation.

B. eliminating the error.

C. by taking more care.

D. by ignoring the least count (if any) of the instrument.

#### SOLUTION

Random errors cannot be eliminated altogether even after taking atmost care. It can only be reduced by taking more number of observations.

Q. 217 The number of significant figure in 2.64  1024 kg are

A. One

B. Two

C. Three

D. Twenty five

#### SOLUTION

Q. 218 An exception in measuring units in terms of length is

A. fermi.

B. micron.

C. debye.

D. light year.

#### SOLUTION

Fermi, micron and light year are the units of length. But debye is the unit of electric dipole moment.

Q. 219 The diameter of a sphere is measured with an instrument having least count 0.001 cm. The diameter is 1.933 cm. The radius to correct significant figures will be

A. 0.9665 cm

B. 0.966 cm

C. 0.967 cm

D. 0.697 cm

#### SOLUTION

Q. 220 One second is defined to be equal to

A. 1650763.73 periods of the Krypton clock

B. 652189.63 periods of the Krypton clock

C. 1650763.73 periods of the Cesium clock

D. 9192631770 periods of the Cesium clock

#### SOLUTION

One second is the duration of 9192631770 periods of radiation that corresponds to unperturbed transition between the two hyperfine levels of the ground state of Cs-133 atom.

Q. 221 Pressure is defined as

A. Momentum per unit area

B. Momentum per unit area per unit time

C. Momentum per unit volume

D. Density per unit volume

#### SOLUTION

Pressure = force/area = (Momentum / time) / area

Q. 222 Three physical quantities having same dimensions are

A. Work, Energy and Torque

B. Work, Energy and Pressure

C. Work, Density and Torque

D. Force, Energy and Torque

#### SOLUTION

Work = [M1L2 T-2] Energy = [M1L2 T-2] Torque = [M1L2 T-2]

Q. 223 1 Bar for pressure is equal to

A. 2.1  105 Pascal

B. 2.01  102 Pascal

C. 2.01  106 Pascal

D. 1.01  105 Pascal

#### SOLUTION

Q. 224 How many electrons would make 1 kg ?

A. 2.1 1030

B. 2.1 10-30

C. 1.1 1030

D. 3.1 1030

#### SOLUTION

Number of electrons = total mass / mass of each electron = 1 kg / 9.11 10-31 kg = 1.1 1030

Q. 225 How many ergs are there in 1 kWh?

A. 3.6 1012 ergs

B. 36 1012 ergs

C. 26 1012 ergs

D. 36 1014 ergs

#### SOLUTION

1 kW h = 103 W h = 103 x 107 erg/s x 60 x 60 s            = 36 x 1012 ergs

Q. 226 The no. of radians in one degree is

A. 1.745  10-2

B. 17.45  10-2

C. 1.24  10-3

D. 17.45  10-5

#### SOLUTION

Q. 227 The number of significant figures in 0.06070 are

A. 5

B. 6

C. 4

D. 3

#### SOLUTION

For a number less than 1, 1. the zeroes on the right and left to the decimal point are non significant (0.06070). 2. the trailing zeroes and the zeroes occurring between two non-zero digits are significant (0.06070). 3. All the non -zero digits are significant (0.06070). Therefore, the number of significant figures in 0.06070 are 4.

Q. 228 1 parsec is equal to

A. 3.261 light year

B. 32.6 light year

C. 4.26 light year

D. 3 light year

#### SOLUTION

1 light year = 9.46 x 1015 m 1 par sec = 3.1 x 1016 m 1 par sec = 3.26 light year

Q. 229 If the Size of protons “P” Size of hydrogen atom “H” and Size of nucleus atom “N”. Then the correct order of length for any given objects,

A.  P>N>H

B. N>H>P

C. P>H>N

D. N>P>H

#### SOLUTION

Since Proton Length=10-15, Length of nucleus atom=10-14, Length of hydrogen atom=10-10

Q. 230 Two quantities A and B have different dimensions.Which mathematical operation may be physically meaningful ?

A. A / B

B. A + B

C. A - B

D. A = B

#### SOLUTION

If two physical quantities have different dimensions then they can not be added or subtracted together. They can be multiplied or divided.

Q. 231 0.99 – 0.989 is

A.   0.001

B.  0.010 × 10–1

C.  0.0001

D.  0.1×10–3

#### SOLUTION

0.99 – 0.989 = 0.001

Q. 232 The dimension of surface tension is

A. [MLT-2]

B. [M-2LT-2]

C. [M-1L2T-2]

D. [M-1LT2]

#### SOLUTION

Surface Tension = force / length = MLT-2 / L = [MLT-2]

Q. 233 A certain body weighs 22.42 gm and has a measured volume of 4.7 cm3. The possible error in the measurement of mass and volume are 0.01 gm and 0.1 cm3. Then maximum error in the density will be

A. 22%

B. 2%

C. 0. 2%

D. 0.02%

#### SOLUTION

Q. 234 The dimension of 1/40 in coulomb's force expression is

A. C2N-1m-2.

B. Nm2C-2.

C. Nm2C2.

D. N-1m2C.

#### SOLUTION

F=(1/40)(q1q2/r2)         [Coulomb's law] When charges(q1 and q2), distance(r) and force(F) are measured in coulomb(C), meter(m) and newton(N) respectively then dimension of 1/40(Proportionality constant)= (Fr2)/(q1q2)                                                       = [Nm2C-2]

Q. 235 Dimensional formula of Gravitational constant(G) is

A. ML-1T-2.

B. M-1L3T-2.

C. M-2L-2T-2.

D. M-1L2T-3.

#### SOLUTION

F=(Gm1m2)/r2               (Gravitation Law)
So, G = Fr2/(m1m2) Dimension formulae of force(F), distance(r) and masses(m1 and m2) are MLT-2, L and M respectively. So, dimension formula of G= (MLT-2)L2/M2

Q. 236 1 peta metre is equal to

A. 1018 m.

B. 1012 m.

C. 1015 m.

D. 1010 m.

#### SOLUTION

Peta means a quadrillion(1015) times of a  unit.
So, 1 peta metre = 1015 m.

Q. 237 A proton has a charge of

A. +1.6  10-19 C.

B. -1.6 1018 C.

C. +1.6  10+18 C.

D. -2.6  10+19 C.

#### SOLUTION

An atom is electrically neutral. It consists of an electron, proton and neutron. Charge on a neutron is zero. Charge on an electron is -1.6  10-19 C. So, charge on a proton is equal in magnitude but is opposite in sign to the charge on an electron.

Q. 238 Meter kelvin is the unit  of

A. Rydberg constant.

B. Wein's constant.

C. gravitation constant.

D. gas constant.

#### SOLUTION

Wein's constant (b) =  T SI unit of wavelength() is metres(m). SI unit of temperature(T) is kelvin(K).

Q. 239 The unit of reactance is

A. mho.

B. ohm.

C. ohm/m.

D. ohm-m.

#### SOLUTION

Reactance is the effective resistance of an a.c circuit. So, unit of reactance = unit of resistance.

Q. 240 The number of significant figures in 0.0607 are

A. 5

B. 6

C. 4

D. 3

#### SOLUTION

When the number is less than 1, the zeros to the left of the non-zero didgt are not significant.

Q. 241 Length is measured in

A. centimetres.

B. kilograms.

C. grams.

D. seconds.

#### SOLUTION

Length is measured in centimetre as it is the fundamental unit of length in C.G.S system of units.

Q. 242 The full form of M.K.S. is

A. metre, kilogram, second.

B. mole, kilogram, second.

C. mole, kelvin, second.

D. metre, kilobyte, second.

#### SOLUTION

The full form of M.K.S. is metre,kilogram, second as per the M.K.S systems of units where metre is fundamental unit of length, kilogram is fundamental unit of mass and second is the fundamental unit of time.

Q. 243 The S.I unit of work is

A. Nm.

B. N/m.

C. N.

D. m.

#### SOLUTION

S.I unit of work is Nm, as work = force x distance. S.I unit of force is Newton (N)and distance is metre(m).

Q. 244 The dimensionless quantity is

A. stress.

B. pressure.

C. power.

D. strain.

#### SOLUTION

Strain = change in length /original length or, Strain = change in volume/ original volume. Here, units get cancelled.

Q. 245 The S.I unit of power is

A. watt.

B. newton.

C. volt.

D. joule.

#### SOLUTION

POwer is derived quantity. Power = work/time = Joule/second = watt.

Q. 246 An atomic clock has an accuracy of 1 part is 10". If two such clocks are operated with precision, then after running for 2500 years these will record a difference of nearly

A. 1sec

B. 8 sec

C. 5 sec

D. 108 sec

#### SOLUTION

Difference 8 sec

Q. 247 The full form of M.K.S.?

A. Meter Kilogram Second

B. Mole Kilogram Second

C. Mole Kelvin Second

D. Meter Kilobyte Second

#### SOLUTION

Q. 248 Length cannot be measured in

A. fermi

B. micron

C. debye

D. light year

#### SOLUTION

Debye is the C.G.S. unit of electric dipole moment. 1 fermi = 1 femtometer = 1 fm = 10-15 m 1 micron = 10-6 m 1 light year = 9.46 x 1015 m

Q. 249 The dimensions of 'a' and 'b' in the formula v = a + bt , where 'v' is the velocity and 't' is the time is

A. [L2T-1] and [LT-2]

B. [L3T-1] and [L2T-2]

C. [L-1T1] and [LT-3]

D. [LT-1] and [LT-2]

#### SOLUTION

v = a + bt As L.H.S represents velocity, every term on R.H.S must represent velocity. Thus, a = v = [LT-1]  and bt = v or b = v/t = LT-1/T = [LT-2].

Q. 250 The number of light years in 1 meter is

A. 1.057  10-16 light year

B. 1.057  10-26 light year

C. 1.057  10-18 light year

D. 2.057  10-16 light year

#### SOLUTION

1 light year = 9.46  1015 m Thus, 1 m = 1 / 9.46  1015  light year = 1.057  10-16 light year.

Q. 251 The unit of permittivity of free space, 0 is

A. Coulomb2/(Newton-metre)2

B. Coulomb/Newton-metre

C. Newton-meter2/Coulomb2

D. Coulomb2/Newton-meter2

#### SOLUTION

Unit of 0 is (Coulomb)2/ Newton-metre2.

Q. 252 1 Chander Shekhar Limit is equal to

A. 8.2 1020 kg

B. 2.8 1030 kg

C. 8.2 1030 kg

D. 8.2 1010 kg

#### SOLUTION

1 chander shekhar limit = 1.4 solar mass = 1.4 2 1030 kg = 2.8 1030 kg.

Q. 253 SI unit of magnetic dipole moment is

A. Am-1.

B. Am2.

C. Am.

D. mA-2.

#### SOLUTION

Magnetic dipole moment(M) = IA SI unit of current(I) = A(ampere).
SI unit of area(A) = m2(square meter)

Q. 254 The distance traveled by light in one year is

A. 12x1025 m

B. 15x1023 m

C. One year

D. 9.46x1015 m

#### SOLUTION

1 light year is the distance traveled by light in one year. 1 light year = (3 x 108) x (365 x 24 x 60 x 60) m = 9.46 x 1015 m

Q. 255  The dimensional formula of pressure is

A. [MLT–2]

B. [ML–1T2]

C. [ML–1T–2]

D. [MLT2]

#### SOLUTION

Q. 256 The dimension of rate of flow is

A.  [ML3T–2]

B. [L3T–1]

C. [M1L3T–2]

D. [L3T–2]

#### SOLUTION

Rate of flow represents volume flowing per second.

Q. 257 The order of distance traveled by sound in air in 3 second is

A. 104

B. 106

C. 103

D. 108

#### SOLUTION

Speed of sound in air = 343 m/s Distance traveled by sound in 3 s = 343 x 3 s = 1029 m ~ 103 m

Q. 258 Which one of the following quantities has not been expressed in proper units ?

A. Coefficient of elasticity: N/m2

B. Surface Tension: N/m

C. Energy: kg m/s

D. Pressure: N/m2

#### SOLUTION

Unit of energy is kg m2/s2.

Q. 259 If an atom of size 10-10 m were enlarged to to the size of the earth ( nearly 107 ), how large would its nucleus be ?Take size of nucleus 10-14 m.

A. 10-3 m

B. 10-4 m

C. 105 m

D. 103 m

#### SOLUTION

Size of the nucleus / size of the atom = x / 107 = 10-14/ 10-10 = 10-4 x = 10-4  107 m.

Q. 260 The order of size of our galaxy is

A. 1030 meter

B. 1040 meter

C. 1020 meter

D. 1033 meter

#### SOLUTION

The diameter of our galaxy is 1020 m.

Q. 261 Which one of the following quantities has not been expressed in proper units?

A. Coeff. Of elasticity: N/m2

B. Surface tension: N/m

C. Energy: Kg m/s

D. Pressure: N/m2

#### SOLUTION

Unit of energy is kg m2/s2.

Q. 262 Which of the following do not have any dimension?

A. Stress

B. Presurre

C. Power

D. Strain

#### SOLUTION

Q. 263 Which of the following has same dimensions as Planck’s constant

A.  Torque

B. Angular momentum

C. Gravitatinaol constant

D. Moment

#### SOLUTION

Angular momentum has same dimensions as Planck’s constant

Q. 264 SI unit of magnetic dipole moment is

A. Am-1

B. Am2

C. TmA-1

D. mA-2

#### SOLUTION

Magnetic dipole moment, M = IA Its unit is given by Am2.

Q. 265 Unit of Force is

A.  Joule

B. m/s

C. m/s2

D. Kgm/s2

#### SOLUTION

From F=m .a
=Unit of mass x unit of Acceleration
=Kgm/s2

Q. 266 Light year” and “Year” both measure in

A. Meter and Second respectively

B. Only Second

C. Only Meter

D. Only Minute

#### SOLUTION

Since Light year is a length and Year is time

Q. 267 The velocity of sound in air is 332 m/s. if the unit of distance is km and unit of time in hour, what would be the value of velocity

A. 1074.5 Km/h

B. 1174 Km/h

C. 1195.2 Km/h

D. 1274.5 Km/h

#### SOLUTION

km/h = 1000/(6060) m/s   then, 332m/s = 332 (18/5)km/h                       = 1195.2 Km/h

Q. 268 The mass of earth is

A. 10-12

B. 1023

C. 1025

D. 1027

#### SOLUTION

Mass of earth can be measured by gravitational methods. M = gR2/ G R = 6.4 x 106 m, G = 6.67 x 10-11 Nm2 kg-2 and g = 9.8 m/s2 M = 6.018 x 1024 kg ~ 1025 kg

Q. 269 Give a method to measure size of the atom using Avogadro’s hypothesis?

#### SOLUTION

According to Avogadro’s theorem, all the atoms in a given specimen of the substance occupy 2/3 rd volume occupied by substance. Consider an atom of mass m and volume V.  If M and N are molecular mass of substance and Avogadro’s numbers respectively, the number of atoms of substance having mass.

Assuming that each atom is sphere of radius r, the volume occupied by atoms in substance

Q. 270 The diameter of a thin wire is measured with screw gauge, the value are 2.04 mm, 2.06 mm, 2.06mm, 2.08mm, 2.07mm, 2.05mm. Find the average diameter of wire, absolute error; mean absolute error, relative error and percentage error.

#### SOLUTION

Average diameter =  D = (2.04 + 2.06 +2.06 + 2.08 + 2.07 + 2.05)/6

or,   D = 2.06

Absolute errors in measurements

D1 = 2.06 – 2.04 = 0.02 mm

D2 = 2.06 – 2.06 = 0

D3 = 2.06 – 2.06 = 0

D4 = 2.06 – 2.08 =  - 0.02 mm

D5 = 2.06 – 2.07 =  - 0.01 mm

D6 = 2.06 – 2.05 = 0.01 mm

Mean absolute error,

Q. 271 Find whether the formula given below is dimensionally correct ,

#### SOLUTION

Therefore formula is dimensionally correct.

Q. 272 Find the  expression for centripetal force if it depends upon mass of the body,speed of the body, and the radius of the circular path.

#### SOLUTION

According to question, Fmavbrc

F= kmavbrc

Writing the dimensions of both the sides, we have

[MLT-2]= [M]a[LT-1]b[L]c

[MLT-2]= [MaLb+cT-b]

comparing the powers ,we have

a = 1

b + c = 1

-b = -2  b = 2

2 + c = 1

c = -1

hence, F = kmv2r-1

F = kmv2/r

the value of k found to be 1 hence

F = mv2/r

Q. 273 Read the assertion and reason carefully to mark the correct option out of the options given below.
Assertion: The displacement of the particles along x-axis is given by the equation x = 6 - 8t + t3. Where displacement x is in metres and time t is in second. In the range 0 < t < 2 the speed of the particle is decreasing.
Reason: In the time interval 0 < t < 2 acceleration is negative.

A. Assertion is true but reason is false.

B. Assertion and reason both are false.

C. Both assertion and reason are true and the reason is the correct explanation of the assertion.

D. Both assertion and reason are true but reason is not the correct explanation of the assertion.

#### SOLUTION

The velocity of the particle, $v = dx dt = d( 6 - 8t + t 3 ) dt = -8 + 3t 2 Equation-1$ The velocity of the particle at different instant of time with in time range. 0 < t < 2 is, $v(0)= -8 ms -1 , v(1)= - 6 ms -1 and v(2)= 0 ms -1$ Thus, between time range 0 < t < 2 velocity is negative. Now, let us find the acceleration $a = dv dt = d( -8 + 3 t 2 ) dt = 6t Equation-2 From equation−2 it is clear that acceleration is positive and increasing with time. The sign acceleration is opposite to the sign of velocity it means the particle is speed of the particle is decreasing.$

Q. 274 Read the assertion and reason carefully to mark the correct option out of the options given below.
Assertion: The displacement-time graph for the object’s motion with negative acceleration is curved downward.
Reason: The displacement of the object is proportional to the square of the time interval.

A. Assertion is true but reason is false.

B. Assertion and reason both are false.

C. Both assertion and reason are true and the reason is the correct explanation of the assertion.

D. Both assertion and reason are true but reason is not the correct explanation of the assertion.

#### SOLUTION

The displacement of the object moving in a straight line with negative acceleration can be represented by using second equation of motion. According to this equation, if initial speed of the ball is zero then displacement of the object is proportion to the square of the time interval, which is the equation of parabola. The shape of the curve also depends on the sign of acceleration. For positive acceleration, the curve is upward whereas for negative acceleration, the curve is down ward.

Q. 275 Read the assertion and reason carefully to mark the correct option out of the options given below.
Assertion: The equation of motion can be applicable only if acceleration is constant and it is in the same or opposite direction to velocity.
Reason: During uniform motion, the acceleration of the object remains constant.

A. Assertion is true but reason is false.

B. Assertion and reason both are false.

C. Both assertion and reason are true and the reason is the correct explanation of the assertion.

D. Both assertion and reason are true but reason is not the correct explanation of the assertion.

#### SOLUTION

Equation of motion can be applied if the acceleration is in the same or opposite direction to that of velocity as well as when it is zero. The uniform motion means the object covers equal distance in equal interval of time so its acceleration is zero.

Q. 276 Read the assertion and reason carefully to mark the correct option out of the options given below.
Assertion: The stopping distance of a fast moving car is more on a wet road than on a dry road.
Reason: The acceleration of the car is more on the wet surface than on a dry surface

A. Assertion is true but reason is false.

B. Assertion and reason both are false.

C. Both assertion and reason are true and the reason is the correct explanation of the assertion.

D. Both assertion and reason are true but reason is not the correct explanation of the assertion.

#### SOLUTION

The stopping distance and stopping time of a car depends on the road conditions and driver’s reaction time. For a fixed initial speed, the distance covered by the car is inversely proportional to acceleration of the car. It takes longer to stop a car on wet road than dry.

Q. 277 Read the assertion and reason carefully to mark the correct option out of the options given below.
Assertion: The velocity of a body approaching towards us is more than the velocity of a body moving away from us.
Reason: When the object is moving on the same track, the relative velocity between both bodies is equal to the difference of the individual velocities of both bodies in respective direction.

A. Assertion is true but reason is false.

B. Assertion and reason both are false.

C. Both assertion and reason are true and the reason is the correct explanation of the assertion.

D. Both assertion and reason are true but reason is not the correct explanation of the assertion.

#### SOLUTION

When two bodies are moving in opposite directions, relative velocity between them is equal to sum of the velocity of bodies. But, if the bodies are moving in same direction, their relative velocity is equal to difference in velocity of the bodies.

Q. 278 Read the assertion and reason carefully to mark the correct option out of the options given below.
Assertion: Crowded areas have reduced vehicle speed limits.
Reason: According to third equation of motion, the final velocity of the object depends on acceleration and the distance over which it acts.

A. Assertion is true but reason is false.

B. Assertion and reason both are false.

C. Both assertion and reason are true and the reason is the correct explanation of the assertion.

D. Both assertion and reason are true but reason is not the correct explanation of the assertion.

#### SOLUTION

According to third equation of motion, for fixed value of acceleration, a car that is going twice as fast doesn’t simply stop in twice the distance after applying breaks. Let initial velocity of the car is v and it is moving with acceleration a and stopsafter time interval t. If the body is covering a distance x then from third equation of kinematics, The stopping distance is equal to the square of the initial velocity. Thus, doubling the initial velocity increases the stopping distance by four times.

Q. 279 Read the assertion and reason carefully to mark the correct option out of the options given below.
Assertion: An object slowing down can have positive acceleration.
Reason: The increase and decrease in speed depends upon the direction of motion of an object.

A. Assertion is true but reason is false.

B. Assertion and reason both are false.

C. Both assertion and reason are true and the reason is the correct explanation of the assertion.

D. Both assertion and reason are true but reason is not the correct explanation of the assertion.

#### SOLUTION

In one dimensional motion, the speed of the object decreases when velocity and acceleration have opposite directions. Positive acceleration means that the change in the velocity always remains in the positive direction. Change in speed is independent of direction of motion of the object. The velocity could be in the positive direction when object slowing down or, the velocity could be in the negative direction when the object speeds up.

Q. 280 Read the assertion and reason carefully to mark the correct option out of the options given below.
Assertion: In the given speed-time graph, speed of the object is negative between time interval 0 h to 3h.
Reason: The object is moving in negative direction and its speed decreases with time.

A. Assertion is true but reason is false.

B. Assertion and reason both are false.

C. Both assertion and reason are true and the reason is the correct explanation of the assertion.

D. Both assertion and reason are true but reason is not the correct explanation of the assertion.

#### SOLUTION

The graph and the slopes are in the positive x-y region throughout the time span considered. The body is decelerating in 0 h to 3 h and after that, body is accelerating till 5 h.

Q. 281 Read the assertion and reason carefully to mark the correct option out of the options given below.
Assertion: A ball is thrown upward and returns to the ground. The displacement of ball is always less than the distance covered by the ball during the whole journey.
Reason: The displacement of the ball moving downward or upwards cannot be equal to the distance.

A. Assertion is true but reason is false.

B. Assertion and reason both are false.

C. Both assertion and reason are true and the reason is the correct explanation of the assertion.

D. Both assertion and reason are true but reason is not the correct explanation of the assertion.

#### SOLUTION

Distance is the actual length of the path covered by the ball and it always keeps on increasing from going upward to coming downwards. However, displacement is equal to the distance between point of projection and the maximum height attained by the ball. As the ball moves downward from the maximum height, ball’s displacement keeps on decreasing and shall become zero once it reaches the point of projection.

Q. 282 Read the assertion and reason carefully to mark the correct option out of the options given below.
Assertion: Uniform motion is a kind of motion in which a body covers equal distances in equal intervals of time.
Reason:Uniform motion is always rectilinear motion.

A. Assertion is true but reason is false.

B. Assertion and reason both are false.

C. Both assertion and reason are true and the reason is the correct explanation of the assertion.

D. Both assertion and reason are true but reason is not the correct explanation of the assertion.

#### SOLUTION

In rectilinear motion body moves in a straight-line path so the body may have uniform motion if it covers equal distance in equal intervals of time during its motion. However, a body moving with uniform motion can travel in any direction. For example, the second hand of a clock moves with uniform speed does not have a rectilinear motion.

Q. 283  The instrument to measure instantaneous speed of an automobile is

A. speedometer.

B. audiometer.

C. odometer.

D. tachometer.

#### SOLUTION

The dashboard of an automobile has an instrument called speedometer, which measures how fast the automobile moves. It is usually calibrated in miles per hour or in kilometre per hour.

Q. 284 A ball is dropped downwards, after 1 second another ball is dropped downwards from the same point. The distance between them after 3 seconds will be

A. 20 m

B. 25 m

C. 50 m

D. 98 m

#### SOLUTION

Q. 285  A stone is dropped from the top of the tower and travels 24.5 m in the last second of its journey .The height of the tower is

A. 72 m.

B. 49 m.

C. 44.1 m.

D. 32 m.

#### SOLUTION

Q. 286  S-t graph of a body is a parabola. From the graph we conclude that the body is moving with

A. zero initial velocity and uniform acceleration.

B. uniform speed.

C. uniform velocity.

D. variable speed.

#### SOLUTION

For uniform velocity or speed s = v t, the graph is a straight line.
For zero initial velocity and uniform acceleration, the equation of motion is .

Q. 287 A ball is thrown from a point with a speed v0 at an angle of projection θ. From the same point and at the same instant a person starts running with a constant speed v0/2 to catch the ball. Then the person will catch the ball at an angle of projection

A. 30°

B. 45°

C. 60°

D. 90°

#### SOLUTION

Q. 288  A man throws balls with the same speed vertically upwards one after the other at an interval of 2 seconds. To have more than two balls in the sky at any time the speed of throw should be (g=9.8 m/s2)

A. 19.6 m/s.

B. any speed less than 19.6 m/s.

C. more than 19.6 m/s.

D.  9.8 m/s.

#### SOLUTION

For two balls in air if one is just thrown and other is at the highest point. Then, the time interval between them is t=2 seconds
Now v=u-g t
0=u- g x 2 so, u=19.6 m/s
Hence, for more than two balls in air, the speed must be greater than 19.6 m/s

Q. 289  Which is the most accurate clock?
A. 19.6 m/s.
B. any speed less than 19.6 m/s.
C. more than 19.6 m/s.
D.  9.8 m/s.

#### SOLUTION

Cesium clock (atomic clock)

Q. 290 Do AU and A0 stand for the same unit? Explain.
A. 19.6 m/s.
B. any speed less than 19.6 m/s.
C. more than 19.6 m/s.
D.  9.8 m/s.

#### SOLUTION

Angstrom is the unit for lengthwhich is equal to 10-10 m. it is normally used to denote very small length like wavelength and astronomical unit is also the unit of length. 1 AU =1.46 X 1011  m and used to denote very large distances like the distances between stars and other celestial bodies.

Q. 291 What do you mean by measurement of a physical quantity?
A. 19.6 m/s.
B. any speed less than 19.6 m/s.
C. more than 19.6 m/s.
D.  9.8 m/s.

#### SOLUTION

Measurement means comparison of physical quantity with its unit to find out how many times unit is contained in the physical quantity. Like length of room is 17 m .It means that 1 metre is contained 17 times in the length of room.

Q. 292 What do you mean by unit? Give the characteristics of unit.
A. 19.6 m/s.
B. any speed less than 19.6 m/s.
C. more than 19.6 m/s.
D.  9.8 m/s.

#### SOLUTION

The standard which is taken as reference to compare the physical quantity is called unit. The characterstics of unit are:
1: It should be well defined.
2: It should be easily accessible.
3: It should be easily reproducible at all places.
4: It should not change with time, place and other physical conditions.

Q. 293 Why other methods to measure the time is replaced by cesium atom clock.
A. 19.6 m/s.
B. any speed less than 19.6 m/s.
C. more than 19.6 m/s.
D.  9.8 m/s.

#### SOLUTION

The is because in cesium atom clock, the measurement of time is 200 times accurate in comparison to other methods of measurement.

Q. 294 What do you understand by fundamental quantities and units?
A. 19.6 m/s.
B. any speed less than 19.6 m/s.
C. more than 19.6 m/s.
D.  9.8 m/s.

#### SOLUTION

The quantities which are independent of each other are said to be fundamental quantities. These are length, mass and time. The units of these quantities are called fundamental units. We have some other fundamental quantities temperature, luminous intensity, current, and amount of substance. The unit of these quantities is Kelvin, candela, ampere and mole respectively.

Besides these, two supplementary units are also defined. It is radian for plane angle and steradian for solid angle.

Q. 295 What are the advantages of SI system of units?
A. 19.6 m/s.
B. any speed less than 19.6 m/s.
C. more than 19.6 m/s.
D.  9.8 m/s.

#### SOLUTION

1: It is coherent system of units.

2: This system of units is internationally accepted.

3: This system of unit is very close to cgs system of unit and can easily be changed to that system

4: In this system all multiples and submultiples can be expressed as power of 10.

5: the total number of units is small.

Q. 296 Metre is well defined in terms of wavelength and time in terms of periods of radiation why?
A. 19.6 m/s.
B. any speed less than 19.6 m/s.
C. more than 19.6 m/s.
D.  9.8 m/s.

#### SOLUTION

Metre is defined in terms of wavelength because wavelength is accurately defined. It is reproducible in any laboratory. It is not affected by any physical condition.  Second is defined in terms of periods of radiation because period of radiation is more accurate than other measurement. Period of oscillation is unaffected by time, place and temperature.

Q. 297 How the distance of nearest star can be determined?
A. 19.6 m/s.
B. any speed less than 19.6 m/s.
C. more than 19.6 m/s.
D.  9.8 m/s.

#### SOLUTION

The distance of nearest star can be determined by parallax method suppose S1 is the nearest star whose distance is to be calculated. Consider some distant star S whose position direction is fixed as earth rotates round the sun. First find the parallax angle 1 between distant and nearest star when earth is at position P. After 6 months the earth comes on the position Q and given angle becomes 2. The total parallax angle subtended by nearby star on the earth’s orbital diameter PQ will be the sum of two angles.

Q. 298 Explain the method to determine the size of moon
A. 19.6 m/s.
B. any speed less than 19.6 m/s.
C. more than 19.6 m/s.
D.  9.8 m/s.

#### SOLUTION

The size of moon can be determined by parallax method. Suppose the diameter of moon is D, which is to be measured. Fix the telescope from a point E on the earth on both the sides of moon to get the value of angle O.The angle is called angular diameter of moon. If d is the distance of moon from earth then by using the formula O =D/d. D can be determined.

Q. 299 Explain echo method to find the distance of moon.
A. 19.6 m/s.
B. any speed less than 19.6 m/s.
C. more than 19.6 m/s.
D.  9.8 m/s.

#### SOLUTION

A LASER beam is the source of intense, monochromatic and unidirectional beam. By sending the laser beam towards moon the distance of moon from earth can be determined. If C is the velocity of laser beam and T is the total time taken by laser beam to go towards moon and to come back then by using the formula S=C*T/2 distance of moon from earth can be calculated. The factor of 2 is coming because time noted for double distance.

Q. 300 Explain SONAR method to find the distance of submarines in sea.
A. 19.6 m/s.
B. any speed less than 19.6 m/s.
C. more than 19.6 m/s.
D.  9.8 m/s.