A spring of force constant 800 N/m has an extension of 5cm. The work done in extending it from 5 cm to 15 cm is

16 J
8 J
32 J
24 J

Solution

Workdone, W=\(\frac{1}{2}k(k_{2}^{2})x_{1}^{2}\)

\(=\frac{1}{2}k\left [ (0.15)^{2}-(0.05)^{2} \right ]\)

\(=\frac{1}{2}\times 800\times 0.02=8j\)

A simple pendulum 1 metre long has a bob of 10 kg. If the pendulum swings from a horizontal position, the K.E. of the bob, at the instant it passes through the lowest position of its path is

89 joule
95 joule
98 joule
85 joule

Solution

In the given (V – T) diagram, what is the relation between pressure P₁ and P₂ ?

P₂ > P₁
P₂ < P₁
P₂ = P₁
Cannot be predicted

Solution

Moment of inertia does not depend upon

distribution of mass
axis of rotation
point of application of force
None of these

If R₁ and R₂ are respectively the filament resistances of a 200 watt bulb and a 100 watt bulb designed to operate on the same voltage

R₁ is two times R₂
R₂ is two times R₁
R₂ is four times R₁
R₁ is four times R₂

Solution

Two liquids of densities d₁ and d₂ are flowing in identical capillary tubes uder the same pressure difference. If t₁ and t₂ are time taken for the flow of equal quantities (mass) of liquids, then the ratio of coefficient of viscosity of liquids must be

\(\frac{d_{1}t_{1}}{d_{2}t_{2}}\)
\(\frac{t_{1}}{t_{2}}\)
\(\frac{d_{1}}{d_{2}}\frac{t_{2}}{t_{1}}\)
\(\sqrt{\frac{d_{1}t_{1}}{d_{2}t_{2}}}\)

Two identical geostationary satellites are moving with equal speeds in the same orbit but their sense of rotation brings them on a collision course. The debris will

fall down
move up
begin to move from east to west in the same orbit
begin to move from west to east in the same orbit

Solution

The total momentum will be zero and hence velocity will be zero just after collisiion. The pull of earth will make it fall down.

Which of the following is not due to total internal reflection?

Working of optical fibre
Difference between apparent and real depth of pond
Mirage on hot summer days
Brilliance of diamond

Solution

Difference between apparent and real depth of a pond is due to the refraction of light, not due to the total internal reflection. Other three phenomena are due to the total internal reflection.

A vessel with water is placed on a weighing pan and it reads600 g. Now a ball of mass 40 g and density 0.80 g cm^-3 is sunk into the water with a pin of negligible volume, as shown in figure keeping it sunk. The weighting pan will show a reading

600 g
550 g
650 g
632 g

Solution

Volume of ball =⁴⁰⁄_0.8=50 cm³

Downthrust on water = 50 g.

Therefore reading is 650 g.

A block of mass m on a rough horizontal surface is acted upon by two forces as shown in figure. For equilibrium of block the coefficient of friction between block and surface is

\(\frac{F_{1}\, +\, F_{2}\,sin\, \theta }{mg\, +\, F_{2}\, cos\, \theta }\)
\(\frac{F_{1}\, cos\, \theta +\, F_{2} }{mg\, -\, F_{2}\, sin\, \theta }\)
\(\frac{F_{1}\, +\, F_{2} \, cos\,\theta }{mg\, +\, F_{2}\, sin\, \theta }\)
\(\frac{F_{1}\,sin\, \theta -\, F_{2}}{mg\, +\, F_{2}\, cos\, \theta }\)

Solution

Here, on resolving force F₂ and applying the concept of equilibrium

N = mg + F₂ cos θ, and f = µN

∴ f=µ [mg + F₂ cos θ ] …(i)

Also f = F₁ + F₂ sin θ …(ii)

From (i) and (ii)

µ [mg + F ₂ cos θ] = F₁ + F₂ sin θ

\(\Rightarrow \mu =\frac{F_{1}\, +\, F_{2}\, sin\, \theta }{mg\, +\, F_{2}\, cos\, \theta }\)

When a biconvex lens of glass having refractive index 1.47 is dipped in a liquid,it acts as a plane sheet of glass. This implies that the liquid must have refractive index

equal to that of glass
less than one
greater than that of glass
less than that of glass

Density of liquid is 16.8 g cm^-3. Its value in the International System of Units is

16.8 kgm^-3
168 kgm^-3
1680 kgm^-3
16800 kgm^-3

Solution

16.8 gcm^-3= 16800 Kgm^-3.

Two stones are thrown from the top of a tower, one straight down with an initial speed u and the second straight up with the same speed u. When the two stones hit the ground,they will have speeds in the ratio

2 : 3
2 : 1
1 : 2
1 : 1

Solution

Use v² – u² = 2aS. In both the cases, (u positive or negative) u² is positive.

Two rods A and B of the same material and length have their radii r₁ and r₂ respectively. When they are rigidly fixed at one end and twisted by the same couple applied at the other end, the ratio \(\left ( \frac{Angle\, of\, twist\, at\, the\, end\, of\, A}{Angle\, of\, twist\, at\, the\, end\, of\, B} \right )\) is

\(r_{1}^{2}/r_{2}^{2}\)
\(r_{1}^{3}/r_{2}^{3}\)
\(r_{2}^{4}/r_{1}^{4}\)
\(r_{1}^{4}/r_{2}^{4}\)

Solution

The tempearture of an isolated black body falls from T₁ to T₂ in time t, then t is (Let c be a constant)

\(t=c\left ( \frac{1}{T_{2}}-\frac{1}{T_{1}} \right )\)
\(t=c\left ( \frac{1}{T_{2}^{2}}-\frac{1}{T_{1}^{2}} \right )\)
\(t=c\left ( \frac{1}{T_{2}^{3}}-\frac{1}{T_{1}^{3}} \right )\)
\(t=c\left ( \frac{1}{T_{2}^{4}}-\frac{1}{T_{1}^{4}} \right )\)

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