IIT-JEE 2010 Paper 1 Offline | JEE Advanced Year Wise Previous Years Questions

IIT-JEE 2010 Paper 1 Offline

Numerical

-0

A piece of ice (heat capacity = 2100 J kg^-1 ^oC^-1 and latent heat = 3.36 $$ \times $$ 10⁵ J kg^-1 ) of mass m grams is at - 5 ^oC at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when the ice-water mixture is in equilibrium, it is found that 1 gm of ice has melted. Assuming there is no other heat exchange in the process, the value of m is

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IIT-JEE 2010 Paper 1 Offline

MCQ (More than One Correct Answer)

-1

A few electric field lines for a system of two charges $${Q_1}$$ and $${Q_2}$$ fixed at two different points on the $$x$$-axis are shown in the figure. These lines suggest that

IIT-JEE 2010 Paper 1 Offline Physics - Electrostatics Question 58 English

IIT-JEE 2010 Paper 1 Offline Physics - Electrostatics Question 58 English

$$\left| {{Q_1}} \right| > \left| {{Q_2}} \right|$$

$$\left| {{Q_1}} \right| < \left| {{Q_2}} \right|$$

at a finite distance to the left of $${{Q_1}}$$ the electric field is zero

at a finite distance to the right of $${{Q_2}}$$ the electric field is zero

IIT-JEE 2010 Paper 1 Offline

MCQ (More than One Correct Answer)

-1

A student uses a simple pendulum of exactly 1m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of 1 sec for this and records 40 seconds for 20 oscillations. For this observation, which of the following statement(s) is (are) true?

Error ΔT in measuring T, the time period, is 0.05 seconds

Error ΔT in measuring T, the time period, is 1 second

Percentage error in the determination of g is 5%

Percentage error in the determination of g is 2.5%

IIT-JEE 2010 Paper 1 Offline

MCQ (Single Correct Answer)

-1

Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, 100, 60 and 40 W bulbs have filament resistances R₁₀₀, R₆₀ and R₄₀ respectively, the relation between these resistances is

$${1 \over {{R_{100}}}} = {1 \over {{R_{40}}}} + {1 \over {{R_{60}}}}$$

$${R_{100}} = {R_{40}} + {R_{60}}$$

$${R_{100}} > {R_{60}} > {R_{40}}$$

$${1 \over {{R_{100}}}} > {1 \over {{R_{60}}}} > {1 \over {{R_{40}}}}$$

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