1
IIT-JEE 2010 Paper 2 Offline
MCQ (Single Correct Answer)
+4
-1
If the distance of the point $$P(1, -2, 1)$$ from the plane $$x+2y-2z$$$$\, = \alpha ,$$ where $$\alpha > 0,$$ is $$5,$$ then the foot of the perpendicular from $$P$$ to the planes is
A
$$\left( {{8 \over 3},{4 \over 3}, - {7 \over 3}} \right)$$
B
$$\left( {{4 \over 3},-{4 \over 3}, {1 \over 3}} \right)$$
C
$$\left( {{1 \over 3},{2 \over 3}, {10 \over 3}} \right)$$
D
$$\left( {{2 \over 3},-{1 \over 3}, {5 \over 3}} \right)$$
2
IIT-JEE 2010 Paper 2 Offline
MCQ (Single Correct Answer)
+4
-0
Match the statement in Column-$$I$$ with the values in Column-$$II$$

$$\,\,\,\,$$ $$\,\,\,\,$$ $$\,\,\,\,$$ Column-$$I$$
(A)$$\,\,\,\,$$ A line from the origin meets the lines $$\,{{x - 2} \over 1} = {{y - 1} \over { - 2}} = {{z + 1} \over 1}$$
and $${{x - {8 \over 3}} \over 2} = {{y + 3} \over { - 1}} = {{z - 1} \over 1}$$ at $$P$$ and $$Q$$ respectively. If length $$PQ=d,$$ then $${d^2}$$ is
(B)$$\,\,\,\,$$ The values of $$x$$ satisfying $${\tan ^{ - 1}}\left( {x + 3} \right) - {\tan ^{ - 1}}\left( {x - 3} \right) = {\sin ^{ - 1}}\left( {{3 \over 5}} \right)$$ are
(C)$$\,\,\,\,$$ Non-zero vectors $$\overrightarrow a ,\overrightarrow b $$ and $$\overrightarrow c \,\,$$ satisfy $$\overrightarrow a \,.\,\overrightarrow b \, = 0.$$
$$\left( {\overrightarrow b - \overrightarrow a } \right).\left( {\overrightarrow b + \overrightarrow c } \right) = 0$$ and $$2\left| {\overrightarrow b + \overrightarrow c } \right| = \left| {\overrightarrow b - \overrightarrow a } \right|.$$
If $$\overrightarrow a = \mu \overrightarrow b + 4\overrightarrow c \,\,,$$ then the possible values of $$\mu $$ are
(D)$$\,\,\,\,$$ Let $$f$$ be the function on $$\left[ { - \pi ,\pi } \right]$$ given by $$f(0)=9$$
and $$f\left( x \right) = \sin \left( {{{9x} \over 2}} \right)/\sin \left( {{x \over 2}} \right)$$ for $$x \ne 0$$
The value of $${2 \over \pi }\int_{ - \pi }^\pi {f\left( x \right)dx} $$ is

$$\,\,\,\,$$ $$\,\,\,\,$$ $$\,\,\,\,$$Column-$$II$$
(p)$$\,\,\,\,$$ $$-4$$
(q)$$\,\,\,\,$$ $$0$$
(r)$$\,\,\,\,$$ $$4$$
(s)$$\,\,\,\,$$ $$5$$
(t)$$\,\,\,\,$$ $$6$$

A
$$\left( A \right) \to t;\,\,\left( B \right) \to p,r;\,\,\left( C \right) \to q,s;\,\,\left( D \right) \to r$$
B
$$\left( A \right) \to r;\,\,\left( B \right) \to p;\,\,\left( C \right) \to q,s;\,\,\left( D \right) \to r$$
C
$$\left( A \right) \to t;\,\,\left( B \right) \to p,r;\,\,\left( C \right) \to q;\,\,\left( D \right) \to r$$
D
$$\left( A \right) \to t;\,\,\left( B \right) \to r;\,\,\left( C \right) \to q,s;\,\,\left( D \right) \to r$$
3
IIT-JEE 2010 Paper 2 Offline
MCQ (Single Correct Answer)
+4
-1
A vernier calipers has 1 mm marks on the main scale. It has 20 equal divisions on the Vernier scale which match with 16 main scale divisions. For this Vernier calipers, the least count is
A
0.02 mm
B
0.05 mm
C
0.1 mm
D
0.2 mm
4
IIT-JEE 2010 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-0.75
When liquid medicine of density $$\rho $$ is to be put in the eye, it is done with the help of a dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop. We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension T when the radius of the drop is R. When the force becomes smaller than the weight of the drop, the drop gets detached from the dropper.

If the radius of the opening of the dropper is $$r$$, the vertical force due to the surface tension on the drop of radius R (assuming $$r$$ << R) is

A
$$2\pi rT$$
B
$$2\pi RT$$
C
$${{2\pi {r^2}T} \over R}$$
D
$${{2\pi {R^2}T} \over r}$$
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