Units & Measurements — Page 9 | JEE Physics

Q81

The vernier scale used for measurement has a positive zero error of 0.2 mm. If while taking a measurement it was noted that '0' on the vernier scale lies between 8.5 cm and 8.6 cm, vernier coincidence is 6, then the correct value of measurement is ___________ cm. (least count = 0.01 cm)

A 8.58 cm

B 8.54 cm

C 8.56 cm

D 8.36 cm

Correct Answer

Option B

Solution

Reading = MSR + VSD $\times$ LC $-$ zero error Reading = 8.5 +

{{(0.1) \times 6} \over {10}} - {{0.2} \over {10}} = 8.54

cm

Q82

If time (t), velocity (v), and angular momentum (l) are taken as the fundamental units. Then the dimension of mass (m) in terms of t, v and l is :

A

[{t^{ - 1}}{v^1}{l^{ - 2}}]

B

[{t^1}{v^2}{l^{ - 1}}]

C

[{t^{ - 2}}{v^{ - 1}}{l^1}]

D

[{t^{ - 1}}{v^{ - 2}}{l^1}]

Correct Answer

Option D

Solution

m \propto {t^a}{v^b}{l^c}

m \propto {[T]^a}{[L{T^{ - 1}}]^b}{[M{L^2}{T^{ - 1}}]^c}

{M^1}{L^0}{T^0} = {M^c}{L^{b + 2c}}{T^{a - b - c}}

comparing powers c = 1, b = $-$ 2, a = $-$ 1

m \propto {t^{ - 1}}{v^{ - 2}}{l^1}

Q83

Assertion A : If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is 5 mm and there are 50 total divisions on circular scale, then least count is 0.001 cm. Reason R : Least Count =

{{Pitch} \over {Total\,divisions\,on\,circular\,scale}}

In the light of the above statements, choose the most appropriate answer from the options given below :

A A is not correct but R is correct.

B Both A and R are correct and R is the correct explanation of A.

C A is correct but R is not correct.

D Both A and R are correct and R is NOT the correct explanation of A.

Correct Answer

Option A

Solution

Least Count =

{{Pitch} \over {Total\,divisions\,on\,circular\,scale}}

In 5 revolution, distance travel, 5 mm In 1 revolution, it will travel 1 mm. So least count =

{1 \over {50}}

= 0.02

Q84

A physical quantity 'y' is represented by the formula

y = {m^2}{r^{ - 4}}{g^x}{l^{ - {3 \over 2}}}

If the percentage errors found in y, m, r, l and g are 18, 1, 0.5, 4 and p respectively, then find the value of x and p.

A 5 and

\pm

2

B 4 and

\pm

3

C

{{16} \over 3}

and

\pm {3 \over 2}

D 8 and

\pm

2

Correct Answer

Option C

Solution

{{\Delta y} \over y} = {{2\Delta m} \over m} + {{4\Delta r} \over r} + {{x\Delta g} \over g} + {3 \over 2}{{\Delta l} \over l}

18 = 2(1) + 4(0.5) + xp + {3 \over 2}(4)

$\Rightarrow$ 8 = xp By checking from options.

x = {{16} \over 3},p = \pm {3 \over 2}

Q85

In a Screw Gauge, fifth division of the circular scale coincides with the reference line when the ratchet is closed. There are 50 divisions on the circular scale, and the main scale moves by 0.5 mm on a complete rotation. For a particular observation the reading on the main scale is 5 mm and the 20th division of the circular scale coincides with reference line. Calculate the true reading.

A 5.00 mm

B 5.25 mm

C 5.15 mm

D 5.20 mm

Correct Answer

Option C

Solution

Least count (L. C.) =

{{0.5} \over {50}}

True reading =

5 + {{0.5} \over {50}} \times 20 - {{0.5} \over {50}} \times 5

= 5 + {{0.5} \over {50}}(15) = 5.15

mm

Q86

If the length of the pendulum in pendulum clock increases by 0.1%, then the error in time per day is :

A 86.4 s

B 4.32 s

C 43.2 s

D 8.64 s

Correct Answer

Option C

Solution

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

{{\Delta T} \over T} = {1 \over 2}{{\Delta l} \over l}

\Delta T = {1 \over 2} \times {{0.1} \over {100}} \times 24 \times 3600

\Delta T = 43.2

Q87

Match with .

List - I		List - II
(a)	${R_H}$ (Rydberg constant)	(i)	$kg\,{m^{ - 1}}{s^{ - 1}}$
(b)	h (Planck's constant)	(ii)	$kg\,{m^2}{s^{ - 1}}$
(c)	${\mu _B}$ (Magnetic field energy density)	(iii)	$\,{m^{ - 1}}$
(d)	$\eta$ (coefficient of viscocity)	(iv)	$kg\,{m^{ - 1}}{s^{ - 2}}$ Choose the most appropriate answer from the options given below :

A (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)

B (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i)

C (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)

D (a)-(iii), (b)-(ii), (c)-(i), (d)-(iv)

Correct Answer

Option B

Solution

SI unit of Rydberg const. = m $-$ 1 SI unit of Plank's const. = kg m2s $-$ 1 SI unit of Magnetic field energy density = kg m $-$ 1s $-$ 2 SI unit of coeff. of viscosity = kg m $-$ 1s $-$ 1

Q88

If force (F), length (L) and time (T) are taken as the fundamental quantities. Then what will be the dimension of density :

A [FL

-

4T2]

B [FL

-

3T2]

C [FL

-

5T2]

D [FL

-

3T3]

Correct Answer

Option A

Solution

Density = [FaLbTc] [ML $-$ 3] = [MaLa+bT $-$ 2aLbTc] [M1L $-$ 3] = [MaLa+bT $-$ 2a+c]

\begin{array}{lll}{a = 1} & ; & {a + b = - 3} & ; & { - 2a + c = 0} \\ {} & {} & {1 + b = - 3} & {} & {c = 2a} \\ {} & {} & {b = - 4} & {} & {c = 2} \end{array}

So, density = [F1L $-$ 4T2]

Q89

Match with .

List - I		List - II
(a)	Torque	(i)	MLT $^{ - 1}$
(b)	Impulse	(ii)	MT $^{ - 2}$
(c)	Tension	(iii)	ML $^{ 2}$ T $^{ - 2}$
(d)	Surface Tension	(iv)	MLT $^{ - 2}$ Choose the most appropriate answer from the option given below :

A (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)

B (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)

C (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)

D (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)

Correct Answer

Option A

Solution

torque $\tau$ $\to$ ML2T $-$ 2 (iii) Impulse I $\Rightarrow$ MLT $-$ 1 (i) Tension force $\Rightarrow$ MLT $-$ 2 (iv) Surface tension $\Rightarrow$ MT $-$ 2 (ii) Option (a)

Q90

Which of the following equations is dimensionally incorrect? Where t = time, h = height, s = surface tension,

\theta

= angle,

\rho

= density, a, r = radius, g = acceleration due to gravity, v = volume, p = pressure, W = work done, T = torque,

\in

= permittivity, E = electric field, J = current density, L = length.

A

v = {{\pi p{a^4}} \over {8\eta L}}

B

h = {{2s\cos \theta } \over {\rho rg}}

C

J = \in {{\partial E} \over {\partial t}}

D

W = \Gamma \theta

Correct Answer

Option A

Solution

(a)

{{\pi p{a^4}} \over {8\eta L}} = {{dv} \over {dt}}

= Volumetric flow rate (Poiseuille's law) (b)

h\rho g = {{2s} \over r}\cos \theta

(c)

\varepsilon \times {1 \over {4\pi {\varepsilon _0}}}{a \over {{r^2}}} \times {1 \over \varepsilon } = {q \over t} \times {1 \over {{r^2}}}

= {1 \over {{L^2}}} = I{L^{ - 2}}

LHS

T = {I \over A} = I{L^{ - 2}}

(d) W = $\tau$ $\theta$ Option (a)