1 A⁰ = 10⁻ ¹⁰ m Atomic volume of 1 mole of hydrogen = Avagadros number × volume of hydrogen molecule = 6.023 × 10²³ × π× (10⁻¹⁰ m)³ = 25.2 × 10⁻⁷ m³ Molar volume = 22.4 L = 22.4 × 10⁻³ m³ Molar volume / Atomic volume = 22.4x10⁻³/25.2x10⁷ = 0.89 × 104 ≈ 104 This ratio is large because actual size ofRead more
1 A⁰ = 10⁻ ¹⁰ m
Atomic volume of 1 mole of hydrogen
= Avagadros number × volume of hydrogen molecule
= 6.023 × 10²³ × π× (10⁻¹⁰ m)³
= 25.2 × 10⁻⁷ m³
Molar volume = 22.4 L = 22.4 × 10⁻³ m³
Molar volume / Atomic volume = 22.4×10⁻³/25.2×10⁷ = 0.89 × 104 ≈ 104
This ratio is large because actual size of gas molecule is negligible in
comparison to the inter molecular separation.
Percentage error in measurement of displacement= 5/200x100 Percentage error in measurement of time = 0.2/20x100 Maximum permissible error = 2.5 + 1 = 3.5%
Percentage error in measurement of displacement= 5/200×100
Percentage error in measurement of time = 0.2/20×100
Maximum permissible error = 2.5 + 1 = 3.5%
How many kg make 1 unified atomic mass unit?
1u = 1.66 × 10⁻²⁷ kg
1u = 1.66 × 10⁻²⁷ kg
See lessOne mole of an ideal gas at STP occupies 22.4 L. What is the ratio of molar volume to atomic volume of a mole of hydrogen? Why is the ratio so large. Take radius of hydrogen molecule to be 1ºA.
1 A⁰ = 10⁻ ¹⁰ m Atomic volume of 1 mole of hydrogen = Avagadros number × volume of hydrogen molecule = 6.023 × 10²³ × π× (10⁻¹⁰ m)³ = 25.2 × 10⁻⁷ m³ Molar volume = 22.4 L = 22.4 × 10⁻³ m³ Molar volume / Atomic volume = 22.4x10⁻³/25.2x10⁷ = 0.89 × 104 ≈ 104 This ratio is large because actual size ofRead more
1 A⁰ = 10⁻ ¹⁰ m
See lessAtomic volume of 1 mole of hydrogen
= Avagadros number × volume of hydrogen molecule
= 6.023 × 10²³ × π× (10⁻¹⁰ m)³
= 25.2 × 10⁻⁷ m³
Molar volume = 22.4 L = 22.4 × 10⁻³ m³
Molar volume / Atomic volume = 22.4×10⁻³/25.2×10⁷ = 0.89 × 104 ≈ 104
This ratio is large because actual size of gas molecule is negligible in
comparison to the inter molecular separation.
The escape velocity v of a body depends on- (i) the acceleration due to gravity ‘g’ of the planet, (ii) the radius R of the planet. Establish dimensionally the relation for the escape velocity.
v a gᵅ Rᵇ ⇒ v = k gᵅ Rᵇ, K → dimensionless proportionality constant [V] = [g]ᵅ [R]ᵇ [M⁰L¹T⁻¹] = [M⁰L¹T⁻²] [M⁰L¹T⁰]ᵇ equating powers 1 = a + b – 1 = – 2a ⇒ a= 1/2 b= 1 - a = 1-1/2 = 1/2 v=k √gR
v a gᵅ Rᵇ ⇒ v = k gᵅ Rᵇ, K → dimensionless proportionality constant
See less[V] = [g]ᵅ [R]ᵇ
[M⁰L¹T⁻¹] = [M⁰L¹T⁻²] [M⁰L¹T⁰]ᵇ
equating powers
1 = a + b
– 1 = – 2a ⇒ a= 1/2
b= 1 – a = 1-1/2 = 1/2
v=k √gR
A calorie is a unit of heat or energy and it equals 4.2 J where 1J = 1 kg m²s⁻². Suppose we employ a system of units in which unit of mass is α kg, unit of length is β m, unit of time γs. What will be magnitude of calorie in terms of this new system.
n₂ = n₁ [M₁/ M₂]ᵅ (L₁/ L₂)ᵇ (T₁/ T₂)ᶜ = 4.2 ( kg/αkg)¹ (m/βm)² (s/γs)⁻² n₂ = 4.2 α^(-1) β⁻² γ⁻²
n₂ = n₁ [M₁/ M₂]ᵅ (L₁/ L₂)ᵇ (T₁/ T₂)ᶜ
See less= 4.2 ( kg/αkg)¹ (m/βm)² (s/γs)⁻²
n₂ = 4.2 α^(-1) β⁻² γ⁻²
A laser light beamed at the moon takes 2.56s and to return after reflection at the moon’s surface. What will be the radius of lunar orbit
t = time taken by laser beam to go to the moon = distance between earth and moon = d = c × t/2 = 3x10⁸x 2.56/2 = 3.84 x 10⁸ m.
t = time taken by laser beam to go to the moon =
See lessdistance between earth and moon
= d = c × t/2
= 3×10⁸x 2.56/2
= 3.84 x 10⁸ m.
When the planet Jupiter is at a distance of 824.7 million km from the earth, its angular diameter is measured to be 35.72′′ of arc. Calculate diameter of Jupiter.
θ = 35.72′′ 1′′ = 4.85 × 10⁻⁶ radian ⇒ 35.72′′ = 35.72 × 4.85 ×10⁻⁶ rad. d = DQ = 824.7 × 35.72 × 4.85 × 10⁻⁶ = 1.4287 × 10⁵ km
θ = 35.72′′
See less1′′ = 4.85 × 10⁻⁶ radian ⇒ 35.72′′ = 35.72 × 4.85 ×10⁻⁶ rad.
d = DQ = 824.7 × 35.72 × 4.85 × 10⁻⁶
= 1.4287 × 10⁵ km
A physical quantity is measured as Q = (2.1 ± 0.5) units. Calculate the percentage error in (1) Q2 (2) 2Q.
P = Q2 ∆p/p= 2∆Q/Q = 2(0.5/2.1) = 1.0/2.1 = 0.476 ∆p/p x 100% = 47.6% = 48% R=2Q ∆R/R=∆Q/Q ⇒ 0.5/2.1 = 0.238 ∆R/R x 100% = 24%
P = Q2
See less∆p/p= 2∆Q/Q = 2(0.5/2.1) = 1.0/2.1 = 0.476
∆p/p x 100% = 47.6% = 48%
R=2Q
∆R/R=∆Q/Q ⇒ 0.5/2.1 = 0.238
∆R/R x 100% = 24%
The length of a rod as measured in an experiment was found to be 2.48m, 2.46m, 2.49m, 2.50m and 2.48m. Find the average length, absolute error and percentage error. Express the result with error limit.
Average length= 2.48+2.46+2.49+2.50+2.48/5 = 12.41/5 = 2.48m Mean absolute error = 0.00+0.02+0.01+0.02+0.00/5 = 0.05 = 0.013m Percentage error = 0.01/2.48x100%=0.04x100% = 0.40% Correct length = (2.48 ± 0.01)m Correct length = (2.48m ± 0.40%)
Average length= 2.48+2.46+2.49+2.50+2.48/5 = 12.41/5 = 2.48m
See lessMean absolute error
= 0.00+0.02+0.01+0.02+0.00/5 = 0.05 = 0.013m
Percentage error = 0.01/2.48×100%=0.04×100%
= 0.40%
Correct length = (2.48 ± 0.01)m
Correct length = (2.48m ± 0.40%)
If displacement of a body s = (200 ± 5) m and time taken by it t = (20 + 0.2) s, then find the percentage error in the calculation of velocity.
Percentage error in measurement of displacement= 5/200x100 Percentage error in measurement of time = 0.2/20x100 Maximum permissible error = 2.5 + 1 = 3.5%
Percentage error in measurement of displacement= 5/200×100
See lessPercentage error in measurement of time = 0.2/20×100
Maximum permissible error = 2.5 + 1 = 3.5%
5.74 g of a substance occupies 1.2 cm³. Express its density to correct significant figures.
Density = Mass/Volume = 5.74/1.2 = 4.783g/cm³ Here least significant figure is 2, so density = 4.8 g/cm³.
Density = Mass/Volume = 5.74/1.2 = 4.783g/cm³
See lessHere least significant figure is 2, so density = 4.8 g/cm³.