1
JEE Advanced 2013 Paper 1 Offline
+4
-1
The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisions equivalent to 2.45 cm. The 24th division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is
A
5.112 cm
B
5.124 cm
C
5.136 cm
D
5.148 cm
2
IIT-JEE 2012 Paper 1 Offline
+4
-1
In the determination of Young's modulus $$\left( {Y = {{4MLg} \over {\pi l{d^2}}}} \right)$$ by using Searle's method, a wire of length L = 2 m and diameter d = 0.5 mm is used. For a load M = 2.5 kg, an extension l = 0.25 mm in the length of the wire is observed. Quantities d and l are measured using a screw gauge and a micrometer, respectively. They have the same pitch of 0.5 mm. The number of divisions on their circular scale is 100. The contributions to the maximum probable error of the Y measurement
A
due to the errors in the measurements of d and l are the same.
B
due to the error in the measurement of d is twice that due to the error in the measurement of l.
C
due to the error in the measurement of l is twice that due to the error in the measurement of d.
D
due to the error in the measurement of d is four times that due to the error in the measurement of l.
3
IIT-JEE 2011 Paper 1 Offline
+3
-0.75
A dense collection of equal number of electrons and positive ions is called neutral plasma. Certain solids containing fixed positive ions surrounded by free electrons can be treated as neutral plasma. Let 'N' be the number density of free electrons, each of mass 'm'. When the electrons are subjected to an electric field, they are displaced relatively away from the heavy positive ions. If the electric field becomes zero, the electrons begin to oscillate about the positive ions with a natural angular frequency '$${\omega _p}$$' which is called the plasma frequency. To sustain the oscillations, a time varying electric field needs to be applied that has an angular frequency $$\omega$$, where a part of the energy is absorbed and a part of it is reflected. As $$\omega$$ approaches $${\omega _p}$$ all the free electrons are set to resonance together and all the energy is reflected. This is the explanation of high reflectivity of metals.

Taking the electronic charge as 'e' and the permittivity as $$'{\varepsilon _0}'$$. Use dimensional analysis to determine the correct expression for $${\omega _p}$$.

A
$$\sqrt {{{Ne} \over {m{\varepsilon _0}}}}$$
B
$$\sqrt {{{m{\varepsilon _0}} \over {Ne}}}$$
C
$$\sqrt {{{N{e^2}} \over {m{\varepsilon _0}}}}$$
D
$$\sqrt {{{m{\varepsilon _0}} \over {N{e^2}}}}$$
4
IIT-JEE 2011 Paper 2 Offline
+3
-0.75
The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the ball has a relative error of 2 %, the relative percentage error in the density is
A
0.9 %
B
2.4 %
C
3.1 %
D
4.2 %
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