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6 months ago

Class 11th Work, Energy and Power

6.26 A 1 kg block situated on a rough incline is connected to a spring of spring constant 100 N m^–1 as shown in Fig. 6.17. The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest.

Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless.

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Payal Gupta

Contributor-Level 10

6.26 Mass of the block = 1 kg

Spring constant = 100 N/m

Displacement of the block, x = 10 cm = 0.1 m

At equilibrium, normal reaction, R = $m g \cos ? 37 °$

Frictional force, F = $? R$ = $m g \sin ? 37 °$

Net force acting on the block down on the incline = $m g \sin ? 37 °$ - F

= $m g \sin ? 37 °$ - $? m g \cos ? 37 °$

=mg ( $\sin ? 37 ° - ? \cos ? 37 °$ )

At equilibrium,

Work done = Potential energy of the stretched string

mg ( $\sin ? 37 ° - ? \cos ? 37 °$ ) = (1/2)kx2

1 x 10 x ( $\sin ? 37 ° - ? \cos ? 37 °$ ) = (1/2) x 100 x 0.1

10 x (0.602 – $? 0.798)$ = 0.5 x 100 x 0.1

$? = 0.128$

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6 months ago

Class 11th Work, Energy and Power

6.25 Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track (Fig. 6.16). Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given $θ_{1}$ = 30 $°$ , $θ_{2}$ = 60 $°$ , and h = 10 m, what are the speeds and times taken by the two stones ?

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Payal Gupta

Contributor-Level 10

6.25 From the law of conservation of energy,

the potential energy at the top = Kinetic energy at the bottom

mgh = (1/2)m $v_{1}^{2}$ ….(1)

and

mgh = (1/2)m $v_{2}^{2}$ ….(2)

$v_{1}$ = $v_{2}$ , Both the stones will reach with the same speed

For stone 1, the force acting on the stone 1 is given by , $F_{1}$ = m $* a_{1}$ = mg $\sin ? ?_{1}$

$a_{1}$ = g $\sin ? ?_{1}$

For stone 2, $a_{2}$ = g $\sin ? ?_{2}$

As $?_{2} > ?_{1}$ , $a_{2}$ > $a_{1}$

From v = u + at, we get t = v/a

Therefore $t_{2}$ < $t_{1}$ Stone 2 will reach faster than stone 1

From the law of conservation of energy

mgh = (1/2) mv²

v = $\sqrt{2 g h}$ = 14 m/s ( Given h = 10 m)

The time taken by two stones given as

$t_{1}$ &

...more

6.25 From the law of conservation of energy,

the potential energy at the top = Kinetic energy at the bottom

mgh = (1/2)m $v_{1}^{2}$ ….(1)

and

mgh = (1/2)m $v_{2}^{2}$ ….(2)

$v_{1}$ = $v_{2}$ , Both the stones will reach with the same speed

For stone 1, the force acting on the stone 1 is given by , $F_{1}$ = m $* a_{1}$ = mg $\sin ? ?_{1}$

$a_{1}$ = g $\sin ? ?_{1}$

For stone 2, $a_{2}$ = g $\sin ? ?_{2}$

As $?_{2} > ?_{1}$ , $a_{2}$ > $a_{1}$

From v = u + at, we get t = v/a

Therefore $t_{2}$ < $t_{1}$ Stone 2 will reach faster than stone 1

From the law of conservation of energy

mgh = (1/2) mv²

v = $\sqrt{2 g h}$ = 14 m/s ( Given h = 10 m)

The time taken by two stones given as

$t_{1}$ = v/ $a_{1}$ = v/g $\sin ? ?_{1}$ = 14/(9.81 $\sin ? 30 °)$ = 2.85 s

$t_{2}$ = v/ $a_{2}$ = v/g $\sin ? ?_{2}$ = 14/(9.81 $\sin ? 60 °)$ = 1.65 s

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6 months ago

Class 11th Work, Energy and Power

6.24 A bullet of mass 0.012 kg and horizontal speed 70 m s–1 strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.

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Payal Gupta

Contributor-Level 10

6.24 Mass of the bullet, $m_{b}$ = 0.012 kg

Initial speed of the bullet, u = 70 m/s

Mass of the wooden block , $m_{w}$ = 0.4 kg

Initial speed of the wooden block = 0

Let's assume, final speed of the bullet = v

Applying the law of conservation of momentum

$m_{b} * u + m_{w} * 0 = (m_{b} + m_{w}) v$

Hence v = ( $m_{b} * u) / (m_{b} + m_{w})$ = 2.04 m/s

Let h be the height by which the block rise. Applying law of conservation of energy

Potential energy of the combined bullet + block = Kinetic energy of the combination

$(m_{b} + m_{w}) * g * h =$ (1/2) $* (m_{b} + m_{w}) * v^{2}$

h = $v^{2}$ /2g = 0.212 m

The heat produced = Initial kinetic energy of the bullet – final kinetic energy of the combination

= (1/2) $m_{b}$ $u^{2}$ - (1/2) $* (m_{b} + m_{w}) * v^{2}$

= (1/2

...more

6.24 Mass of the bullet, $m_{b}$ = 0.012 kg

Initial speed of the bullet, u = 70 m/s

Mass of the wooden block , $m_{w}$ = 0.4 kg

Initial speed of the wooden block = 0

Let's assume, final speed of the bullet = v

Applying the law of conservation of momentum

$m_{b} * u + m_{w} * 0 = (m_{b} + m_{w}) v$

Hence v = ( $m_{b} * u) / (m_{b} + m_{w})$ = 2.04 m/s

Let h be the height by which the block rise. Applying law of conservation of energy

Potential energy of the combined bullet + block = Kinetic energy of the combination

$(m_{b} + m_{w}) * g * h =$ (1/2) $* (m_{b} + m_{w}) * v^{2}$

h = $v^{2}$ /2g = 0.212 m

The heat produced = Initial kinetic energy of the bullet – final kinetic energy of the combination

= (1/2) $m_{b}$ $u^{2}$ - (1/2) $* (m_{b} + m_{w}) * v^{2}$

= (1/2) $*$ 0.012 X $70^{2}$ - (1/2) $* (0.012 + 0.4) * {2.04}^{2}$ J

= 29.4 – 0.857 J

= 28.54 J

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6 months ago

Class 11th Work, Energy and Power

6.23 A family uses 8 kW of power.

(a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW?

(b) Compare this area to that of the roof of a typical house.

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Payal Gupta

Contributor-Level 10

6.23 Power used by the family = 8 kW = 8000 W

(a) Solar energy received = 200 W/ $m^{2}$

Percentage conversion of Solar energy to Electrical energy = 20%

If the area required is A then 0.2 $* A * 200 = 8000$

A = 200 $m^{2}$ . The comparable roof size is 14.14 X 14.14 m

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6 months ago

Class 11th Hydrogen

9.6. Complete the following reactions.

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Vishal Baghel

Contributor-Level 10

Answer: (i) 3H₂? (g)+2MoO_3?? Mo₂? O₃? +3H_2?O (l)

(ii) CO (g) + H₂ (g)? CH₃OH

(iii)C₃H₈ (g) + 3H₂O (g)? 3CO + 7H₂ (g)

(iv) Zn (s) + NaOH (aq)? Na₂ZnO₂ (s) + H₂ (g)

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6 months ago

Class 11th Motion in a Plane

Q.4.21 A particle starts from the origin at t = 0 s with a velocity of 10.0 ? m/s and moves in the x-y plane with a constant acceleration of (8.0 ? + 2.0 ?) m s^-2.

(a) At what time is the x- coordinates of the particle 16 m? What is the y-coordinate of the particle at that time?

(b) What is the speed of the particle at the time?

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Vishal Baghel

Contributor-Level 10

Ans.4.21:

(a) Velocity , $\overset{?}{v}$ = 10.0 ? m/s

Acceleration, $\overset{?}{a}$ = (8.0 ? + 2.0 ?) m s-2

We know $\overset{?}{a}$ = $\frac{d \overset{?}{v}}{d t}$ = 8.0 ? + 2.0 ?

$\overset{?}{d v}$ = (8.0 ? + 2.0 ?)dt

Integrating both sides we get $\overset{?}{v}$ (t) = 8.0t ? + 2.0t ? + $\overset{?}{u}$ ,

Where, $\overset{?}{u} =$ velocity vector of the particle at t =0

$\overset{?}{v} =$ velocity vector of the particle at time t

But $\overset{?}{v}$ = $\frac{\underset{d r}{?}}{d t}$

$\overset{?}{d r}$ = $\overset{?}{v}$ dt

= (8.0t ? + 2.0t ? + $\overset{?}{u}$ )dt

Integrating both sides with the condition at t = 0, r =0 and at t =t, r = r

$\overset{?}{r} =$ $\overset{?}{u}$ t + ½ $*$ 8.0 t2 ? + ½ $*$ 2.0 $*$ t2 ? = $\overset{?}{u}$ t + 4.0 t2 ? + t2 ?

Substituting the value of $\overset{?}{u}$ , we get

$\overset{?}{r} =$ ( 10.0 ?)t + 4.0 t2 ? + t2 ? . This equation can be expressed as

x ? + y ? = 4.0 t2 ? + ( 10.0t + t2) ?

Since the motion of the particle is confined to the x-y plane, on equating the coefficients of ? and ?, we get

x = 4.0 t2 and y = 10.0t + t2

t = $\sqrt{x / 4}$

(a) When x = 16m, t = 2 s, y = 24m

(b) Velocity of the particle

$\overset{?}{v}$ (t) = 8.0t ? + 2.0t ? + $\overset{?}{u}$

At t = 2 s,

$\overset{?}{v}$ (t) = 8.0 $*$ 2 ? + 2.0 $* 2$ ? + 10 $*$ ? = 16 ? + 14 ?

The magnitude of $\overset{?}{v}$ (t) is given by

$|\overset{?}{v}|$ = ( 162 + 142)1/2 = 21.26 m/s

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6 months ago

Class 11th Work, Energy and Power

6.22 A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated.

(a) How much work does she do against the gravitational force ?

(b) Fat supplies 3.8 * 10⁷J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?

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Payal Gupta

Contributor-Level 10

6.22 Mass lifted, m = 10 kg

Height to which the mass lifted, h = 0.5 m

No of repetitions, n = 1000

(a) Work done against gravitational force,

W = nmgh = 1000 $* 10 * 9.81 * 0.5 =$ 49050 J

(b) Mechanical energy supplied by 1 kg fat, with 20% efficiency rate = 0.2 $*$ 3.8 $* 10^{7}$ = 0.76 $* 10^{7}$ J/kg

Fat used by dieter = 49050 / (0.76 $* 10^{7})$ kg = $6.45 * 10^{- 3}$ kg

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6 months ago

Class 11th Work, Energy and Power

6.21 The blades of a windmill sweep out a circle of area A.

(a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t ?

(b) What is the kinetic energy of the air ?

(c) Assume that the windmill converts 25% of the wind's energy into electrical energy, and that A = 30 m2, v = 36 km/h and the density of air is 1.2 kg m–3. What is the electrical power produced ?

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Payal Gupta

Contributor-Level 10

6.21 Given, the area of the windmill sweep = A, Wind velocity = v

The volume of air passing through the blade = Av

Let the density of air be $?$ , the mass of air passing through the blade = $? A v$

(a) The mass of air passing through the blade in time t = $? A v t$

(b) The kinetic energy of air = $(\frac{1}{2}) m v^{2}$ = $(\frac{1}{2}) ? A v t v^{2}$ = $(? A t v^{3})$ /2 …. (1)

(c) Area, A = 30 $m^{2}$ , v = 36 km/h = 10 m/s, density of air be $?$ = 1.2 kg/ $m^{3}$

Total wind energy, from eqn. (1) = 18 kW

Electrical energy = 25 % of wind energy = 0.25 $* 18 = 4.5 k W$

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6 months ago

Class 11th Work, Energy and Power

6.20 A body of mass 0.5 kg travels in a straight line with velocity v =a x ^3/2 where a = 5 m^–1/2s–1. What is the work done by the net force during its displacement from x = 0 to x = 2 m?

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Payal Gupta

Contributor-Level 10

6.20 Mass of the body = 0.5 kg

Velocity, v = a x 3/2

a = 5 m–1/2 s–1

At x =0, the initial velocity, u = 0

At x = 2, the final velocity, v = 5 $* 2^{3 / 2}$ = 14.142 m/s

Work done by the system = increase in K.E. of the body = (1/2)m ( $v^{2}$ - $u^{2}$ )

= (1/2) $* 0.5 *$ 14.142 $* 14.142$ = 50 J

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6 months ago

Class 11th Work, Energy and Power

6.19 A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg s–1. What is the speed of the trolley after the entire sand bag is empty?

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Payal Gupta

Contributor-Level 10

6.19 Given, the mass of the trolley, $m_{t}$ = 300 kg, mass of the sand bag, $m_{s}$ = 25 kg, uniform velocity of the trolley, v = 27 kmph = 0.75 m/s

Since there is no external force acting on the system, the speed of the trolley will remain unchanged even after entire sand is empty. 27 kmph is the answer.

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