1
JEE Advanced 2013 Paper 1 Offline
MCQ (Single Correct Answer)
+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
2
IIT-JEE 2012 Paper 1 Offline
MCQ (Single Correct Answer)
+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
3
IIT-JEE 2011 Paper 1 Offline
MCQ (Single Correct Answer)
+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 $${\omega _p}$$.
4
IIT-JEE 2011 Paper 2 Offline
MCQ (Single Correct Answer)
+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
Questions Asked from Units & Measurements (MCQ (Single Correct Answer))
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JEE Advanced Subjects
Physics
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Laws of Motion
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Properties of Matter
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