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Published on: 07/06/2021
QB365 provides detailed and simple solution for every Book back Questions in class 11 Physics Subject. It will helps to get more idea about question pattern in every book back questions with solution.
Download Tamil Nadu 11th Standard Physics question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
Questions + Answers key
Take MCQ Physics Test1.
A capacitor of capacitance C = 3.0 ± 0.1\(\mu \)F is charged to a voltage of V = 18 ± 0.4 Volt. Calculate the charge Q [Use Q = CV].
2.
The force F acting on a body moving in a circular path depends on mass of the body (m), velocity (v) and radius (r) of the circular path. Obtain the expression for the force by dimensional analysis method. (Take the value of k = 1)
3.
The parallal x of a heavenly body measured from two points diametrically opposite on equator of earth is 2'. Calculate the distance of the heavenly body. [Given radius of the earth = 6400 km] [1" = 4.85\(\times\)10-6 rad]
4.
What do you mean by propagation of errors? Explain the propagation of errors in addition and multiplication.
5.
Check the correctness of the equation\(\frac { 1 }{ 2 } \)mv2 = mgh using dimensional analysis method.
1.
\((C+\triangle C)=(3.0\pm0.1)\mu F\)
\((V+\triangle V)=(18\pm 0.4)V\)
Q = CV
\(Q=3.0\times10^{-6}\times18=54\times10^{-6}\) coulomb
\(\text {Error in } C=\frac{\Delta C}{C} \times 100=\frac{0.1}{3} \times 100=3.3 \%\)
Error in \(V=\frac{\triangle V}{V}\times100=\frac{0.4}{18}\times100=2.2\%\)
Error in Q = Error in C + Error in V
= 3.3% + 2.2% = 5.5%
\(\therefore\) Charge Q = (54\(\times\)106 土 5.5%)C
2.
F ∝ ma vb rc ;
F = k ma vb rc
where k is a dimensionless constant of proportionality. Rewriting above equation in terms of dimensions and taking k= 1, we have
[MLT-2] = [M]a [LT-1]b [L]c
= [MaLbT-bLc]
[MLT-2] = [MaLb+cT-b]
Comparing the powers of M, L and T on both sides
a = 1 ; b + c = 1; -b =-2
2 + c = 1 ; b = 2;
a = 1 b = 2 and c = -1
From the above equation we get
F = mavbrc
F =m1v2r-1
or F = \(\frac { m{ v }^{ 2 } }{ r } \)
3.
Angle \(\theta\) = 2' = 2\(\times\)60" = 120" = 120\(\times\)4.85\(\times\)10-6 rad
\(\theta\) = 5.82\(\times\)10-4 rad;
d = 2 x r
d = 2 x 6400 = 12800 x 103 m
The distance of heavenly body
\(D=\frac{d}{\theta}=\frac{12800\times10^3}{5.82\times10^{-4}}\)
D = 2.19\(\times\)1010m.
4.
Propagation of errors
A number of measured quantities may be involved in the final calculation of an experiment. Different types of instruments might have been used for taking readings. Then we may have to look at the errors in measuring various quantities, collectively.
The error in the final result depends on
(i) The errors in the individual measurements
(ii) On the nature of mathematical operations performed to get the final result. So we should know the rules to combine the errors.
The various possibilities of the propagation or combination of errors in different mathematical operations are discussed below:
(i) Error in the sum of two quantities:
Let A\(\triangle\) and \(\triangle\)B be the absolute errors in the two quantities A and B respectively. Then,
Measured value of A = A \(\pm\triangle\) A
Measured value of B = B \(\pm\triangle\) B
Consider the sum, Z = A + B
The error \(\triangle\) Z in Z is the given by
Z \(\pm\triangle\) Z = (A \(\pm\triangle\)A) + ( B \(\pm\triangle\) B)
= ( A + B ) \(\pm\) (\(\triangle\)A+ \(\triangle\) B)
= Z \(\pm\) ( \(\triangle\) A + \(\triangle\) B )
(or) \(\triangle\)Z = \(\triangle\) A+ \(\triangle\) B
The maximum possible error in the sum of two quantities is equal to the sum of the absolute errors in the individual quantities.
(ii) Error in the difference of two quantities:
Let \(\triangle\)A and \(\triangle\)B be the absolute errors in the two quantities, A and B, respectively. Consider the product Z = AB
Let \(\triangle\)A and \(\triangle\)B be the absolute errors in the two quantities, A and B, respectively. Consider the product Z = AB
The error \(\triangle\)Z in Z is given by \(Z \pm \Delta Z=(A \pm \Delta A)(B \pm \Delta B)\)
\(=(A B) \pm(A \Delta) \pm(B \Delta A) \pm(\Delta A . \Delta B)\)
Dividing L.H.S by Z and R.H.S by AB, we get,
\(1 \pm \frac{\Delta Z}{Z} \cdot 1 \pm \frac{\Delta B}{B} \pm \frac{\Delta A}{A} \pm \frac{\Delta A}{A} \cdot \frac{\Delta B}{B}\)
As \(\triangle\)A/A, \(\triangle\)B/B are both small quantities, their product term \(\frac{\Delta A}{A} \cdot \frac{\Delta B}{B}\) can be neglected. The maximum fractional error in Z is
\(\frac{\Delta Z}{Z}=\pm\left(\frac{\Delta A}{A}+\frac{\Delta B}{B}\right)\)
The maximum error in difference of two quantities is equal to the sum of the absolute errors in the individual quantities.
(iii) Error in the division or quotient of two quantities
Let \(\triangle\)A and \(\triangle\)B be the absolute errors in the two quantities A and B respectively.
Consider the quotient, \(\mathrm{Z}=\frac{A}{B}\)
The error \(\triangle\)Z in Z is given by
\(Z \pm \Delta Z =\frac{A \pm \Delta A}{B \pm \Delta B}=\frac{A\left(1 \pm \frac{\Delta A}{A}\right)}{B\left(1 \pm \frac{\Delta B}{B}\right)} \)
\(=\frac{A}{B}\left(1 \pm \frac{\Delta A}{A}\right)\left(1 \pm \frac{\Delta B}{B}\right)^{-1}\)
or \(Z \pm \Delta Z=Z\left(1 \pm \frac{\Delta A}{A}\right)\left(1 \pm \frac{\Delta B}{B}\right)\)
[using (1+x) n \(\approx\) 1+n x, when x<1]
Dividing both sides by Z, we get,
\(1 \pm \frac{\Delta Z}{Z} =\left(1 \pm \frac{\Delta A}{A}\right)\left(1 \pm \frac{\Delta B}{B}\right)
\)
\(=1 \pm \frac{\Delta Z}{Z} \pm \frac{\Delta B}{B} \pm \frac{\Delta A}{A} \cdot \frac{\Delta B}{B}\)
As the terms \(\triangle\)A /A and \(\triangle\)B/ B are small, their product term can be neglected. The maximum fractional error in Z is given by
\(\frac{\Delta Z}{Z}=\left(\frac{\Delta A}{A}+\frac{\Delta B}{B}\right)\)
The maximum fractional error in the quotient of two quantities is equal to the sum of their individual fractional errors.
5.
Dimension formula for
\(\frac { 1 }{ 2 } \)mv2 = [M][LT-1]2 = [ML2T-2]
Dimension formula for
mgh = [M][LT-2][L] = [ML2T-2]
[ML2T-2] = [ML2T-2]
Both sides are dimensionally the same, hence the equations \(\frac { 1 }{ 2 } \)mv2 = mgh is dimensionally correct
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