Units & Measurements

Q: The smallest division on the main scale of a Vernier calipers is 0.1 cm . Ten division of the vernier scale correspond to nine division of main scale . The figure below on the left shows the reading of this calipers with no gap between its two jaws . The figure on the right shows the reading with a solid sphere . The figure on the right shows the reading with a solid sphere held between the jaws . The correct diameter of the sphere is

(a)3.07 cm (b) 3.11 cm (iii) 3.15 cm (iv) 3.17 cm

Ans: (c)

Solution: Least Count $= (1 – \frac{9}{10})0.1 = 0.01 cm $

Zero Error = -0.1 + 0.06 = -0.04 cm

Final Reading = 3.1 + 0.01 x 1 = 3.11 cm

So , Correct measurement = 3.11 + 0.04 = 3.15 cm

Q: Sometimes it is convenient to construct a system of units so that all quantities can be expressed in terms of only one physical quantity . In one such system , dimensions of different quantities are given in terms of a quantity X as follows : [position] = [X^α] ; [speed] = [X^β] ; [acceleration] = [X^p] ; [linear momentum] = [X^q] ; [force] = [X^r] . Then
(a) α + p = 2 β

(b) p + q -r = β

(c) p-q + r = α

(d) p + q + r = β

Ans: (a,b)

Solution: position/speed = time

[v] = X^β

[a] = X^p

Speed/acceleration = time

linear momentum = X^q

F = X^r

Momentum/Force = time

Position/speed = speed/acceleration = momentum/force

$\frac{X^{\alpha}}{X^{\beta}} = \frac{X^{\beta}}{X^p} = \frac{X^q}{X^r}$

$X^{\alpha – \beta} = X^{\beta – p} = X^{q – r}$

$ \alpha – \beta = \beta – p = q – r$

α + p = 2 β

p + q -r = β

Q: The density of a solid metal sphere is determined by measuring its mass and its diameter . The maximum error in the density of the sphere is (x/100) % . If the relative error in measuring the mass and the diameter are 6.0 % and 1.5 % respectively , the value of x is – – – –

Sol: Density , $\displaystyle \rho = \frac{M}{V}$

$\displaystyle \rho = \frac{M}{\frac{4}{3}\pi R^3}$

$\displaystyle \rho = \frac{M}{\frac{4}{3}\pi (D/2)^3}$

$\displaystyle \rho = \frac{M}{\frac{\pi}{6}D^3}$

$\displaystyle \frac{\Delta \rho}{\rho} \times 100 = (\frac{\Delta M}{M}\times 100 ) + 3 (\frac{\Delta D}{D}\times 100 ) $

= 6 % + 3(1.5 %) = 10.5 %

(x/100) % = 10.5 %

x = 1050

Q: A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings 5.50 mm , 5.55 mm , 5.54 mm , 5.65 mm . The average of these four readings is 5.5375 mm and standard deviation of the data is 0.07395 mm . The average diameter of the pencil should therefore be recorded as

(a) (5.5375 ± 0.0739 ) mm

(b) (5.5375 ± 0.0740 ) mm

(c) (5.538 ± 0.074 ) mm

(d) (5.54 ± 0.07 ) mm

Ans: (d)

Sol: Here Average diameter d_avg = 5.5375 mm ≈5.54 mm

Standard deviation Δd = 0.07395 mm ≈0.07 mm

Average diameter of the pencil = (5.54 ± 0.07 ) mm

Q: A quantity f is given by $\displaystyle f = \sqrt{\frac{h c^5}{G}} $ , where c is speed of light , G universal gravitational constant and h is planck’s constant . Dimension of f is that of

(a) Area

(b) Volume

(c) Momentum

(d) Energy

Ans: (d)

Sol: $\displaystyle h = E/\nu = [M L^2 T^{-1}]$ , $\displaystyle G = \frac{F r^2}{m_1 m_2} = [M^{-1} L^3 T^{-2}]$

$\displaystyle f = \sqrt{\frac{h c^5}{G}} $

$\displaystyle f = [\frac{[ML^2 T^{-1}][LT^{-1}]^5}{M^{-1}L^3T^{-2}}]^{1/2} $

$\displaystyle f = [M^2 L^4 T^{-4}]^{1/2} $

$\displaystyle f = [M L^2 T^{-2}] $

This is the dimension of Energy .

Q: The dimension of stopping potential V_o in Photoelectric effect in units of Planck’s constant ‘h’ , speed of light ‘c’ and Gravitational constant ‘G’ and ampere ‘A’ is

(a) $h^{-2/3} c^{-1/3}G^{4/3}A^{-1}$

(b) $h^{2/3} c^{5/3}G^{1/3}A^{-1}$

(c) $h^2 c^{1/3}G^{3/2}A^{-1}$

(d) $h^{1/3} c^{-1/3}G^{2/3}A^{-1}$

Ans: None of these .

Sol: $\displaystyle V_o = h^x c^y G^z A^w $

V_o = W/q , E = h ν

$\displaystyle \frac{[ML^2 T^{-2}]}{[AT]} = [M L^2 T^{-1}]^x [LT^{-1}]^y [M^{-1}L^3 T^{-2}]^z [A]^w$

$\displaystyle [M L^2 T^{-3}A^{-1}] = [M^{x-z}] [L^{2x+y+3z}][T^{-x-y-2z}A^w] $

On equating Powers ,

x-y =1

2x + y + 3z = 2

-x-y-2z = -3

w = -1

Q: It is found that $\displaystyle |\vec{A} + \vec{B} | = |\vec{A}| $ . This necessarily implies.

(a) $\displaystyle \vec{B} = 0 $

(b) $\displaystyle \vec{A} ,\vec{B} $ are antiparallel

(c) $\displaystyle \vec{A} ,\vec{B} $ are perpendicular

(d) $\displaystyle \vec{A} ,\vec{B} $ ≤ 0

Ans: (a) , (b)

Sol: $\displaystyle |\vec{A} + \vec{B} |^2 = |\vec{A}|^2 $

$\displaystyle (\vec{A} + \vec{B} ).(\vec{A} + \vec{B} ) = \vec{A}.\vec{A} $

$\displaystyle A^2 + B^2 + 2 \vec{A}.\vec{B} = A^2 $

$\displaystyle B^2 + 2 \vec{A}.\vec{B} = 0 $

Q: Which of the following are not a unit of time ?

(a) Second

(b) Parsec

(c) Year

(d) Light year

Ans: (b) & (d)

Parsec and Light year are the units of distance

1 Parsec = 3.1 x 10¹⁶ m

1 light year = 9.46 x 10¹⁵ m

These are not unit of time . These are unit of distance .

Q: If P,Q,R are physical quantities, having different dimensions, which of the following combinations can never be a meaningful quantity?

(a) (P-Q)/R

(b) PQ-R

(c) PQ/R

(d) (R-Q)/P

Click to See Answer :

Ans: (a) and (d)

Physical quantity having same dimension can be added or subtracted .

 

Q: The mean length of an object is 5 cm. which of the following measurements is most accurate ?

(a) 4.9 cm

(b) 4.805 cm

(c) 5.25 cm

(d) 5.4 cm

Click to See Answer :

Ans: (a)

4.9 can be easily round off to 5 , hence option (a) is the most accurate