Units & Measurements — Page 6 | JEE Physics

Q51

In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s. The length of pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of g is close to :-

A 0.2%

B 3.5%

C 0.7%

D 6.8%

Correct Answer

Option D

Solution

Time period of a pendulum (T) =

2\pi \sqrt {{l \over g}}

$\Rightarrow$ T2 =

4{\pi ^2}{l \over g}

$\Rightarrow$

g = {{4{\pi ^2}l} \over {{T^2}}}

Fractional change

\left( {{{dg} \over g}} \right) \times 100 = \left( {{{dl} \over l}} \right) \times 100 - \left( {2{{dT} \over T}} \right) \times 100

$\therefore$ Maximum possible percentage error,

\left( {{{dg} \over g}} \right) \times 100 = \left( {{{dl} \over l}} \right) \times 100 + \left( {2{{dT} \over T}} \right) \times 100

Error in time period(dT) = least count of time = 1 second and T = 30 second Error in length(dl) = least count of length = 1 mm and

l

= 55.0 cm $\therefore$

\left( {{{dg} \over g}} \right) \times 100 =

\left( {{{0.1} \over {55}}} \right) \times 100 + 2\left( {{1 \over {30}}} \right) \times 100

= 6.8%

Q52

The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5

\mu

m diameter of a wire is :

A 500

B 100

C 200

D 50

Correct Answer

Option C

Solution

Least count =

{{Pitch} \over {Number\,\,of\,\,division\,\,on\,\,circular\,scale}}

5 $\times$ 10 $-$ 6 =

{{{{10}^{ - 3}}} \over N}

N = 200

Q53

If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young,s modulus will be:

A V

-

2A2F2

B V

-

4A

-

2F

C V

-

4A2F

D V

-

2A2F

-

2

Correct Answer

Option C

Solution

We know, Young's modulus (Y) =

{{{F \over A}} \over {{{\Delta l} \over l}}}

$\therefore$ [Y] =

{{\left[ {ML{T^{ - 2}}} \right]} \over {\left[ {{L^2}} \right]}}

= [ ML-1T-2] Let [Y] = [V]x [A]y [F]z $\therefore$ [ ML-1T-2] = [LT-1]x [LT-2]y [MLT-2]z $\Rightarrow$ [ ML-1T-2] = [ Mz Lx + y + z T-x -2y - 2z For dimensional balance, the dimension on both sides should be same.

So, z = 1 x + y + z = -1 $\Rightarrow$ x + y = -2 ........(1) and -x -2y - 2z = -2 $\Rightarrow$ x + 2y = 0 ...........(

2) By solving those two equations we get, x = -4 and y = 2 $\therefore$ [Y] = V $-$ 4A2F1

Q54

The diameter and height of a cylinder are measured by a meter scale to be 12.6

\pm

0.1 cm and 34.2

\pm

0.1 cm, respectively. What will be the value of its volume in appropriate significant figures ?

A 4264.4

\pm

81.0 cm3

B 4264

\pm

81 cm3

C 4300

\pm

80 cm3

D 4260

\pm

80 cm3

Correct Answer

Option D

Solution

Volume of cylinder(V) = $\pi$ r2h =

\pi {{{d^2}} \over 4}h

=

3.14 \times {{{{\left( {12.6} \right)}^2}} \over 4} \times 34.2

= 4260

{{\Delta V} \over V} = 2{{\Delta d} \over d} + {{\Delta h} \over h} = 2\left( {{{0.1} \over {12.6}}} \right) + {{0.1} \over {34.2}}

= 0.0188 $\therefore$

\Delta

V = 0.0188 $\times$ 4260 = 80

Q55

If E, L, M and G denote the quantities as energy, angular momentum, mass and constant of gravitation respectively, then the dimensions of P in the formula P = EL2M

-

5G

-

2 are :

A [M0 L1 T0]

B [M

-

1 L

-

1 T2]

C [M1 L1 T

-

2]

D [M0 L0 T0]

Correct Answer

Option D

Solution

E = ML2T $-$ 2 L = ML2T $-$ 1 m = M G = M $-$ 1L+3T $-$ 2 P =

{{E{L^2}} \over {{M^5}{G^2}}}

[P] =

{{(M{L^2}{T^{ - 2}})({M^2}{L^4}{T^{ - 2}})} \over {{M^5}({M^{ - 2}}{L^6}{T^{ - 4}})}} = {M^0}{L^0}{T^0}

Q56

The density of a material in SI units is 128 kg m–3 . In certain units in which the unit of length is 25 cm and the unit of mass is 50 g, the numerical value of density of the material is -

A 40

B 640

C 16

D 410

Correct Answer

Option A

Solution

Here given that density of a material in SI units is 128 kg m–3 And a new unit system is introduced where 1 unit of length = 25 cm and 1 unit of mass = 50 g You should know that, physical quantity is same in any unit system.

And to calculate a physical quantity yo should know two things (1) numerical value of the physical quantity (n) (2) unit of the physical quantity (u) And n $\times$ u = constant in any unit system.

Here in SI unit system, n1 = 128 u1 = kg/m3 And in new unit system, n2 = ?

u2 = 50gm/(25cm)3 As n1u1 = n2u2 $\therefore$ 128 $\times$ (kg/m3) = n2 $\times$ 50gm/(25cm)3 $\Rightarrow$ 128 $\times$

{{1000\,gm} \over {{{\left( {100\,cm} \right)}^3}}}

= n2 $\times$

{{50\,gm} \over {{{\left( {25\,cm} \right)}^3}}}

$\Rightarrow$ n2 = 128 $\times$

{{20} \over {{4^3}}}

= 40

Q57

Amount of solar energy received on the earth’s surface per unit area per unit time is defined a solar constant. Dimension of solar constant is :

A MLT–2

B ML0T–3

C M2L0T–1

D ML2T–2

Correct Answer

Option B

Solution

Solar constant =

{E \over {AT}}

=

{{\left[ {{M^1}{L^2}{T^{ - 2}}} \right]} \over {\left[ {{L^2}T} \right]}}

= [ ML0T–3 ]

Q58

Dimensional formula for thermal conductivity is (here K denotes the temperature):

A MLT–3K–1

B MLT–2K–2

C MLT–2K

D MLT–3K

Correct Answer

Option A

Solution

$\therefore$

{{d\theta } \over {dt}} = kA{{dT} \over {dx}}

$\Rightarrow$ k =

{{\left( {{{d\theta } \over {dt}}} \right)} \over {A\left( {{{dT} \over {dx}}} \right)}}

$\Rightarrow$ [k] =

{{\left[ {M{L^2}{T^{ - 3}}} \right]} \over {\left[ {{L^2}} \right]\left[ {K{L^{ - 1}}} \right]}}

= [MLT–3K–1]

Q59

A quantity x is given by

\left( {{{IF{v^2}} \over {W{L^4}}}} \right)

in terms of moment of inertia I, force F, velocity v, work W and Length L. The dimensional formula for x is same as that of :

A Coefficient of viscosity

B Force constant

C Energy density

D Planck's constant

Correct Answer

Option C

Solution

x =

\left( {{{IF{v^2}} \over {W{L^4}}}} \right)

$\therefore$ [x] =

{{\left[ {M{L^2}} \right]\left[ {ML{T^{ - 2}}} \right]{{\left[ {L{T^{ - 1}}} \right]}^2}} \over {\left[ {M{L^2}{T^{ - 2}}} \right]{{\left[ L \right]}^4}}}

= [ML-1T-2] = [Energy density]

Q60

A physical quantity z depends on four observables a, b, c and d, as z =

{{{a^2}{b^{{2 \over 3}}}} \over {\sqrt c {d^3}}}

. The percentages of error in the measurement of a, b, c and d are 2%, 1.5%, 4% and 2.5% respectively. The percentage of error in z is :

A 13.5 %

B 14.5%

C 16.5%

D 12.25%

Correct Answer

Option B

Solution

z =

{{{a^2}{b^{{2 \over 3}}}} \over {\sqrt c {d^3}}}

$\Rightarrow$

{{dz} \over z} \times 100 = \left( {2{{da} \over a} + {2 \over 3}{{db} \over b} + {1 \over 2}{{dc} \over c} + 3{{d\left( d \right)} \over d}} \right) \times 100

% error in z =

\left( {2 \times 2 + {2 \over 3} \times 1.5 + {1 \over 2} \times 4 + 3 \times 2.5} \right)

% = ( 4 + 1 + 2 + 7.5 ) % = 14.5 %