JEE Advanced 2013 Paper 2 Offline | JEE Advanced Year Wise Previous Years Questions

JEE Advanced 2013 Paper 2 Offline

MCQ (Single Correct Answer)

-0.75

A small block of mass 1 kg is released from rest at the top of a rough track. The track is circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure, below, is 150 J. (Take the acceleration due to gravity, g = 10 m/s²)

JEE Advanced 2013 Paper 2 Offline Physics - Work Power & Energy Question 19 English Comprehension

The magnitude of the normal reaction that acts on the block at the point Q is

7.5 N

8.6 N

11.5 N

22.5 N

JEE Advanced 2013 Paper 2 Offline

MCQ (Single Correct Answer)

-0.75

JEE Advanced 2013 Paper 2 Offline Physics - Work Power & Energy Question 20 English Comprehension

The speed of the block when it reaches the point Q is

5 ms^-1

10 ms^-1

$$10\sqrt 3 $$ ms^-1

20 ms^-1

JEE Advanced 2013 Paper 2 Offline

MCQ (Single Correct Answer)

-0.75

Match List I with List II and select the correct answer using the codes given below the lists:

List I

P. Boltzmann Constant
Q. Coefficient of viscosity
R. Plank Constant
S. Thermal conductivity

List II

1. [ML²T^-1]
2. [ML^-1T^-1]
3. [MLT^-3K^-1]
4. [ML²T^-2K^-1]

$$\eqalign{ & P\,\,\,\,Q\,\,\,\,R\,\,\,\,S \cr & 3\,\,\,\,\,\,1\,\,\,\,\,2\,\,\,\,\,4 \cr} $$

$$\eqalign{ & P\,\,\,\,Q\,\,\,\,R\,\,\,\,S \cr & 3\,\,\,\,\,\,2\,\,\,\,\,1\,\,\,\,\,4 \cr} $$

$$\eqalign{ & P\,\,\,\,Q\,\,\,\,R\,\,\,\,S \cr & 4\,\,\,\,\,\,2\,\,\,\,\,1\,\,\,\,\,3 \cr} $$

$$\eqalign{ & P\,\,\,\,Q\,\,\,\,R\,\,\,\,S \cr & 4\,\,\,\,\,\,1\,\,\,\,\,2\,\,\,\,\,3 \cr} $$

JEE Advanced 2013 Paper 2 Offline

MCQ (More than One Correct Answer)

-0.75

Using the expression $$2d\sin \theta = \lambda $$, one calculates the values of d by measuring the corresponding angles $$\theta $$ in the range 0 to 90^o. The wavelength $$\lambda $$ is exactly known and the error in $$\theta $$ is constant for all values of $$\theta $$. As $$\theta $$ increases from 0^o

the absolute error in d remains constant

the absolute error in d increases

the fractional error in d remains constant

the fractional error in d decreases

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