1 Bq = one disintegration per second
1 curie = 3.7 x 10¹⁰ disintegration per second
                 (Bq)
\(\therefore 1Bq=\frac { 1 }{ 3.7\times { 10 }^{ 10 } } \)Curie
= 2.703 x 10^-11ci

Initial temperature, t₁ = 0°C
Final temperature, t₂ = 25°C
Coefficient of linear expansion of metal,
α₁ =18 x 10^-6 C^-1
Measured length, l₁ = 50 cm
To find: True (or) Actual length l₂ = ?
\(\frac { \triangle l }{ l\triangle T } =\frac { ({ l }_{ 2 }-{ l }_{ 1 }) }{ { l }_{ 1 }(t_{ 2 }-{ t }_{ 1 }) } \)
l₂ = I₁ (1 + α (t₂ - t₁)
∴ l₂ = 50 x (1 + 18 x 10 ^-6 x (25 - 0)
= 50 x (1 + 450 x 10^-6)
= 50 x (1 + 0.00045)
∴ l₂ = 50.225 cm
The true length of the object at 0⁰C =50.225 cm.

Object distance, u =25 cm
virtual image distance, v =-10 cm
To find: Magnification, m = ?
m = \(\frac{\text { Distance of the image }}{\text { Distance of the object }}=\frac{v}{u}\)
= \(\frac { -10 }{ 25 } \) = -0.4
Magnification,
m = -0.4

Time, t = 10 s
Mass of body, m = 10 kg
Initial velocity of the body, u₁ = 0
Distance, d = 500 cm
= 500 x 10^-2 m
To find : Force, F = ?
F=m₁\(\frac { ({ u }_{ 1 }-{ v }_{ 1 }) }{ t } \)

v₁=\(\frac { distance }{ time\quad taken } \)
F=v₁=\(\frac { 500\times 10^{ -2 } }{ 10 } \)
Force, F=0.5 N

Impulse = Change in momentum
P = 60 N ; t = 10 s
Change in momentum = Force x time
= 60 x 10
Change in momentum = 600 Ns

Given :
Potential difference, V = 110 V
Charge, q = 5 C
To find: Work done, W = ?
Solution
\(V=\frac { W }{ q } \therefore q\times V=W\)
W= 5 x 110 = 550 J
Work, W = 550J

Mass = 100g = 0.1 kg;
Initial speed u = 20 ms^-1
Final velocity v = 0
Change in momentum = ?
mv - mu = 0.1 (0 - 20)
Change in momentum = -2 kg ms^-1.

Number of moles = \(\frac{Mass}{Atomic \ mass}\)
∴ Mass = Number of moles x Atomic mass
= 0.5 x 16
= 8g.

Let T° C be the required temperature. Let v₁ and v₂ be the velocity of sound at temperatures T₁ K and T₂ K respectively. T₁ = 273 K (0°C) and T₂ = (T°C + 273) K
\(\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}}=\sqrt{\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}}=\sqrt{\frac{273+\mathrm{T}}{273}}=2\)
Here, it is given that, v₂ / v₁ = 2.
So, 273 + T / 273 = 4
T = (273 x 4) - 273
= 819° C.

pOH = -log₁₀ [OH^-]
pOH = -log₁₀ [1 x 10^-9]
pOH = -(log₁₀ 1.0 + log₁₀ 10^-9)
pOH = -(0-9 log₁₀10)
pOH = -(0-9)
pOH = 9.