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 ….

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

IIT

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

Click to See Solution :
Ans: (c)

Sol: 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

 

Sometimes it is convenient to construct a system of units so that all quantities can be expressed in terms of only one physical quantity ….

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] = [Xα] ; [speed] = [Xβ] ; [acceleration] = [Xp] ; [linear momentum] = [Xq] ; [force] = [Xr] . Then
(a) α + p = 2 β

(b) p + q -r = β

(c) p-q + r = α

(d) p + q + r = β

Click to See Solution :
Ans: (a,b)

Sol: position/speed = time

[v] = Xβ

[a] = Xp

Speed/acceleration = time

linear momentum = Xq

F = Xr

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

 

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 ….

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

A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings ….

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 davg = 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

A quantity f is given by f = √(hc^5/G) , where c is speed of light , G universal gravitational constant and h is planck’s constant …

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 .