pH + pOH = 14
pOH = 14 – 4.5 = 9.5
pOH = 9.5

Formula: The moment of a force M = F × d
Given: F = 40 N and d = 90 cm = 0.9 m.
Hence, moment of the force = 40 × 0.9 = 36 N m.

KOH is a strong base and dissolve in water and gives
\(\mathrm{KOH_{(aq)}} \rightarrow \mathrm{K}^{+}_{(s)}+\mathrm{OH}^{-}_{(aq)}\)
Each KOH molecules gives one OH^- ion. So 1.0 x 10^-5 molar molar solution of KOH gives 1.0 x 10^-5  OH^- ions.
\({\left[\mathrm{OH}^{-}\right] } =1.0 \times 10^{-5}  \)
\(\mathrm{pOH} =-\log _{10}\left[\mathrm{OH}^{-}\right]  \)
\(=-\log _{10} 1.0 \times 10^{-5}  \)
\(=-(-5) \log _{10} 10  \)
\(\mathrm{pOH} =5 \times 1  \)
\(\mathrm{pOH} =5  \)
\(\mathrm{pH}+\mathrm{pOH} =14  \)
\(\mathrm{pH} =14-\mathrm{pOH} \)
=14 - 5 
pH = 9
The pH of 1.0 x10^-5 molar solution of KOH is 9.

Sulphuric acid dissociates in water as:
H₂SO_4(aq) → 2 H⁺_(aq) + SO₄^2-_(aq)
Each mole of sulphuric acid gives two mole of H⁺ ions in the solution. One litre of H₂SO₄ solution contains 5 × 10^–5 moles of H₂SO₄ which would give 2 × 5 × 10^–5 = 10 × 10^–5 or 1.0 × 10^–4 moles of H⁺ ion in one litre of the solution.
Therefore,
[H⁺] = 1.0 × 10^–4 mol litre^–1
pH = –log₁₀[H⁺] = –log₁₀10^–4 = –(–4 × log₁₀10)
= –(–4 × 1) = 4

1 Bq = one disintegration per second
one curie = 3.7 × 10¹⁰ Bq
1 Bq = \(\frac { 1 }{ 3.7\times { 10 }^{ 10 } } \) curie
∴ 3.7 x 10³ G Bq = 3.7 x 10³ x 10⁹ x \(\frac { 1 }{ 3.7\times { 10 }^{ 10 } } \)
= 100 curie

molecular mass of H₂SO₄
= (1 × 2) + (32 × 1) + (16 × 4)
= 2 + 32 + 64
= 98 g
% of S in H₂SO₄ = \(\frac { Mass\ of\ sulphur }{ Moleculer\ mass\ of\ H_2\ SO_4 } \times 100\)
% of S in H₂SO₄ = \(\cfrac { 32 }{ 98 } \times 100\)
= 32.65%

Mass percentage (w / w) =  \(\frac{\text { Mass of the solute }}{\text { Mass of the solution }} \times 100\)
\(10=\frac{\text { Mass of the urea }}{500} \times 100\)
Mass of urea = 50 g

Mass of the solute = 25 g
Mass of the solvent = 100 g
\(\begin{aligned} \text { Mass Percentage } \end{aligned}=\frac{\text { Mass of the solute }}{\text { Mass of the solution }} \times 100\)
\(\begin{aligned} \text { Mass Percentage } \end{aligned}=\frac{\text { Mass of the solute }}{\begin{array}{l} \text { Mass of the solute }+ \text { Mass of the solvent } \end{array}} \times 100\)
\(=\frac {25}{25+100}\times\)100
= \(\frac{25}{125}\times\)100
= 20%

Mass of potassium chloride in 100 g of water in saturated solution = 37 g
Mass of potassium chloride in 60 g of water in saturated solution = \(\frac{37}{100}\)× 60
= 22.2 g

A) Since the resistors are connected in parallel, the potential difference across each resistor is same (i.e. V=10V)
Therefore, the current through R₁ is,
\({ I }_{ 1 }=\frac { V }{ { R }_{ 1 } } =\frac { 10 }{ 5 } =2A\)
Current through \({ R }_{ 2 }={ I }_{ 2 }=\frac { V }{ { R }_{ 2 } } =\frac { 10 }{ 10 } =1A\)
Current through \({ R }_{ 3 }={ I }_{ 3 }=\frac { V }{ { R }_{ 3 } } =\frac { 10 }{ 20 } =0.5A\)
B) Total current in the circuit, I = I₁ + I₂ + I₃
= 2 + 1 + 0.5 = 3.5 A
C) Total resistance in the circuit  \(\frac { 1 }{ { R }_{ P } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } +\frac { 1 }{ { R }_{ 3 } } \)
\(=\frac { 1 }{ 5 } +\frac { 1 }{ 10 } +\frac { 1 }{ 20 } \)
\(=\frac { 4+2+1 }{ 20 } \)
\(\frac { 1 }{ { R }_{ P } } =\frac { 7 }{ 20 } \)
Hence, \({ R }_{ P }=\frac { 20 }{ 7 } =2.857\Omega \) 

When the source is moving towards the stationary listener, the expression for apparent frequency is
\(n'=\left( \frac { v }{ v-{ v }_{ s } } \right) n\)
\(1000=\left( \frac { 330 }{ 330-50 } \right) \times n\)
\(n=\left( \frac { 1000 \times 280 }{ 330 } \right)\)
n = 848.48 Hz
The actual frequency of the sound is 848.48 Hz. When the source is moving away from the stationary listener, the expression for apparent frequency is
\(n'=\left( \frac { v }{ v+{ v }_{ s } } \right) n\)
=\(\left( \frac { 330 }{ 330+50 } \right) \times 848.48\)
= 736.84 Hz

f = -0.3m, v = -0.2m
\(\cfrac { 1 }{ f } =\cfrac { 1 }{ v } -\cfrac { 1 }{ u } \)
\(\cfrac { 1 }{ u } =\cfrac { 1 }{ v } -\cfrac { 1 }{ f } \)
\(\cfrac { 1 }{ u } =\cfrac { 1 }{ -0.2 } -\cfrac { 1 }{ -0.3 } =\cfrac { -10 }{ 6 } \)
\(u=\cfrac { -6 }{ 10 } =-0.6m\)

When the source is moving towards the stationary listener, the expression for apparent frequency is
\(n'=\left( \frac { v }{ v-{ v }_{ s } } \right) n\)
= \(\left( \frac { v }{ v-\left( \frac { 1 }{ 10 } \right) v } \right) n=\left( \frac { 10 }{ 9 } \right) n\)
= \(\left( \frac { 10 }{ 9 } \right) \times 90=100\) Hz
n'=100 Hz

Data:
Initial pressure (P₁)= P
Final Pressure (P₂) = 4P
Initial volume (V₁) = 20cc = 20cm³
Final volume (V₂) = ?
Using Boyle's Law, PV = constant
P₁V₁ = P₂V₂
\({ V }_{ 2 }=\frac { { P }_{ 1 } }{ { P }_{ 2 } } \times { v }_{ 1 }\)
\(=\frac { P }{ 4P } \times 20{ cm }^{ 3 }\)
V₂ = 5 cm³

1)  0.3 mole of aluminium (Atomic mass of Al = 27)
\(\text { Number of moles }=\frac{\text { Mass of } \mathrm{Al}}{\text { Atomic mass of } \mathrm{Al}}\)
Mass = No. of moles × atomic mass
So, mass of Al = 0.3 × 27
= 8.1 g
2) 2.24 litre of SO₂ gas at S.T.P
 Molecular mass of SO₂ = 32 + (16 × 2)
= 32 + 32 = 64
\(\text { Number of moles of } \mathrm{SO}_2=\frac{\begin{array}{c} \text { Given volume of } \mathrm{SO}_2 \text { at S.T.P } \end{array}}{\begin{array}{c} \text { Molar volume } \mathrm{SO}_2 \text { at S.T.P } \end{array}}\)
Number of moles of SO₂ = \(\frac{2.24}{22.4}\)
= 0.1 mole
\(\text { Number of moles }=\frac{\text { Mass }}{\text { Molecular mass }}\)
Mass = No. of moles × molecular mass
Mass = 0.1 × 64
Mass of SO₂ = 6.4 g
iii) 1.51 × 10²³ molecules of water
Molecular mass of H₂O = 18
\(\text { Number of moles }=\frac{\begin{array}{c} \text { Number of molecules of }\text { water } \end{array}}{\text { Avogadro's number }}\)
= 1.51 × 10²³ / 6.023 × 10²³
= 1 / 4
= 0.25 mole
\(\text { Number of moles }=\frac{\text { Mass }}{\text { Molecular mass }}\)
0.25 = mass / 18
Mass = 0.25 × 18
Mass = 4.5 g
iv) 5 × 10²³ molecules of glucose
Molecular mass of glucose = 180
\(\text { Mass of glucose }=\frac{\begin{array}{c} \text { Molecular mass } \times \text { number of particles } \end{array}}{\text { Avogadro's number }}\)
= (180 × 5 × 10²³) / 6.023 × 10²³
= 149.43 g