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 .

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

Q: The dimension of stopping potential Vo 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 $

Vo = 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

It is found that |A + B|= |A|. This necessarily implies

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

Click to See Answer :
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 $