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Answers
Question 1 : The measure mass and volume of a body are 22.42 g and 4.7 cm³ respectively with possible errors of 0.01 g and 0.1 cm³. The maximum error in density is :
Answer 1 :
Given :
m = 22.42 g
v = 4.7 cm³
∆m = 0.01 g
∆v = 0.1 cm³
We know that, when two quantities are multiplied or divided, the relative error in the result is the sum of the relative errors in the multipliers.
Hence, ∆R/R = ∆m/m + ∆v/v
Let's find the relative error of mass :
⇒ 0.01/22.42
⇒ 0.00044 g
Now, finding the relative error of velocity :
⇒ 0.1/4.7
⇒ 0.02 cm³
Therefore, ∆R/R = ∆m/m + ∆v/v
⇒ ∆R/R = 0.00044 + 0.02
⇒ ∆R/R = 0.02044
Finding the percentage error in density : 0.02044 × 100
⇒ 2.044 % = 2%
Option B = 2%
Question 2 : If a physical quantity is represented by x = [M^aL^bT^-c] and if the percentage error in the measurement of M, L and T are α%, β% and γ% respectively, then the total percentage error in x is :
Answer 2 :
Here, the quantities M and L are in multiplication and the quantity T is divided.
Hence, the final percentage error = (∆x/x)% = a(∆M/M)% + b(∆L/L)% + c(∆T/T)%
We are already given that the percentage error of M, L and T are α%, β% and γ% respectively.
Hence, (∆x/x)% = (aα + bβ + cγ)%
Option A = (aα + bβ + cγ)%
Question 3 : If the units of force, energy and velocity are 10 N, 100 J and 5 m/s; the units of length, mass and time will be :
Answer 3 :
The Dimensional Formula of :
Force = [MLT^-2]
Energy = [ML²T^-2]
Velocity = [LT^-1]
Given,
F = [MLT^-2] = 10 N
E = [ML²T^-2] = 100 J
V = [LT^-1] = 5 m/s
Dividing Energy by Force, [ML²T^-2]/[MLT^-2] = [L]
Therefore, 100/10 = 10 m.
Now, V = [LT^-1] = 5 m/s = 10*T^-1 = 5 m/s
⇒ 10/5 = T
⇒ T = 2 s
At last, F = [MLT^-2] = 10 N
⇒ M*10*2^-2 = 10
⇒ M = (10 × 2²)/10
⇒ M = 4 kg
Hence, Option (B) = Units of length, mass and time will be : 10 m, 2 s and 4 kg respectively.
Question 4 : The least count of a stopwatch is 0.01 s. The time period of 100 oscillations of the pendulum is 100 s. The percentage error in the time period will be :Answer 4 :
Number of oscillations = 100
Time taken = 100 s
The error in 100 oscillations = 0.01 s
So, Time period for 100 oscillations = 100 s ± 0.01 s
Hence, Percentage error = ∆t/t × 100 = 0.01/100 × 100
⇒ Percentage error = 0.01 %
Hence, Option A = 0.01%.
1)
Given :
Mass =m=2..42g
Volume=v=4.7cm3
Δm=0.01g
Δv=0.1cm3
Density (ρ)=mass/volume
Δρ/ρ=Δm/m+Δv/v
=0.01/2.42+0.1/4.7
=0.004+0.02=0.024
%error is Δρ/ρ x100
=0.024x100
=2.4% ≈nearly 2% ∴
maximum error in density is 2%
or
______________________________
Question 1 : The measure mass and volume of a body are 22.42 g and 4.7 cm³ respectively with possible errors of 0.01 g and 0.1 cm³. The maximum error in density is :
Answer 1 :
Given :
m = 22.42 g
v = 4.7 cm³
∆m = 0.01 g
∆v = 0.1 cm³
We know that, when two quantities are multiplied or divided, the relative error in the result is the sum of the relative errors in the multipliers.
Hence, ∆R/R = ∆m/m + ∆v/v
Let's find the relative error of mass :
⇒ 0.01/22.42
⇒ 0.00044 g
Now, finding the relative error of velocity :
⇒ 0.1/4.7
⇒ 0.02 cm³
Therefore, ∆R/R = ∆m/m + ∆v/v
⇒ ∆R/R = 0.00044 + 0.02
⇒ ∆R/R = 0.02044
Finding the percentage error in density : 0.02044 × 100
⇒ 2.044 % = 2%
Option B = 2%
Question 2 : If a physical quantity is represented by x = [M^aL^bT^-c] and if the percentage error in the measurement of M, L and T are α%, β% and γ% respectively, then the total percentage error in x is :
Answer 2 :
Here, the quantities M and L are in multiplication and the quantity T is divided.
Hence, the final percentage error = (∆x/x)% = a(∆M/M)% + b(∆L/L)% + c(∆T/T)%
We are already given that the percentage error of M, L and T are α%, β% and γ% respectively.
Hence, (∆x/x)% = (aα + bβ + cγ)%
Option A = (aα + bβ + cγ)%
Question 3 : If the units of force, energy and velocity are 10 N, 100 J and 5 m/s; the units of length, mass and time will be :
Answer 3 :
The Dimensional Formula of :
Force = [MLT^-2]
Energy = [ML²T^-2]
Velocity = [LT^-1]
Given,
F = [MLT^-2] = 10 N
E = [ML²T^-2] = 100 J
V = [LT^-1] = 5 m/s
Dividing Energy by Force, [ML²T^-2]/[MLT^-2] = [L]
Therefore, 100/10 = 10 m.
Now, V = [LT^-1] = 5 m/s = 10*T^-1 = 5 m/s
⇒ 10/5 = T
⇒ T = 2 s
At last, F = [MLT^-2] = 10 N
⇒ M*10*2^-2 = 10
⇒ M = (10 × 2²)/10
⇒ M = 4 kg
Hence, Option (B) = Units of length, mass and time will be : 10 m, 2 s and 4 kg respectively.
Question 4 : The least count of a stopwatch is 0.01 s. The time period of 100 oscillations of the pendulum is 100 s. The percentage error in the time period will be :Answer 4 :
Number of oscillations = 100
Time taken = 100 s
The error in 100 oscillations = 0.01 s
So, Time period for 100 oscillations = 100 s ± 0.01 s
Hence, Percentage error = ∆t/t × 100 = 0.01/100 × 100
⇒ Percentage error = 0.01 %
Hence, Option A = 0.01%.