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8 months ago

Class 11th Motion in a Plane

4.15 The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m s^-1 can go without hitting the ceiling of the hall?

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Vishal Baghel

Contributor-Level 10

4.15 The speed of the ball, u = 40 m/s

Ceiling height, h = 25m

Since the ball is thrown, it will follow the path of a projectile. For projectile motion, the maximum height for a body projected at an angle $θ$ is given by

h= $\frac{u^{2} {s i n}^{2} θ}{2 g}$ , sin2 $θ$ = (25 $* 2 * 9.807) / (40 * 40)$ , $θ = 33.61 °$

Horizontal range is given by,=

R = $\frac{u^{2}}{g} s i n 2 θ$ = ((40 $* 40$ ) $* \sin ? 2 * 33.61 °$ )/9.81 = 150.38m

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8 months ago

Class 11th Kinetic Theory

13.3 Figure 13.8 shows plot of PV/T versus P for 1.00*10–3 kg of oxygen gas at two different temperatures.

(a) What does the dotted plot signify?

(b) Which is true: T₁ > T₂ or T₁ < T₂?

(c) What is the value of PV/T where the curves meet on the y-axis?

(d) If we obtained similar plots for 1.00*10^–3 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen

Yields the same value of PV/T (for low pressure high temperature region of the plot) ? (Molecular mass of H₂ = 2.02 u, of O₂ = 32.0 u, R = 8.31 J mo1^–1 K^–1.)

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Payal Gupta

Contributor-Level 10

13.3 (a) The dotted plot in the graph signifies the ideal behaviour of the gas, i.e. $\frac{P V}{T}$ = $?$ R, (where $?$ is the number of moles and R is the universal gas constant) is a constant quantity. It is not dependent on the pressure of the gas. The curve of the gas at temperature $T_{1}$ is closer to the dotted plot than the curve of the gas at temperature $T_{2 .}$

(b) A real gas approaches the behaviour of an ideal gas when its temperature increases. Therefore $T_{1}$ > $T_{2}$ is true for the given plot.

(c) The value of the ratio $\frac{P V}{T}$ , where the two curves meet, is $?$ R. This is because the ideal gas e

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(b) A real gas approaches the behaviour of an ideal gas when its temperature increases. Therefore $T_{1}$ > $T_{2}$ is true for the given plot.

(c) The value of the ratio $\frac{P V}{T}$ , where the two curves meet, is $?$ R. This is because the ideal gas equation is given as $\frac{P V}{T}$ = $?$ R. From the given data, we have

Molecular mass of oxygen = 32 g

Mass of oxygen = 1 $* 10^{- 3}$ = 1 g

R = 8.314 J/mole/K

Then $\frac{P V}{T}$ = $\frac{1}{32} * 8.314$ = 0.26 J/k

Hence, the value of the ratio $\frac{P V}{T}$ , where the curves meet on the y-axis, is 0.26 J/K.

(d) If we obtain similar plots for 1.00 $* 10^{- 3}$ kg of hydrogen, then we will not get the same values of $\frac{P V}{T}$ at the point where the curves meet on the y-axis. This is because the molecular mass of hydrogen (2.02 u) is different from the oxygen (32.0 u). We have:

$\frac{P V}{T} = 0.26 J / K$ .

Given, molecular mass (M) of $H_{2}$ = 2.02 u

$\frac{P V}{T}$ = $?$ R at constant temperature, where $?$ = $\frac{m}{M}$

M = mass of $H_{2}$

Then, m = $\frac{P V}{T} * \frac{M}{R}$ = $\frac{0.26 * 2.02}{8.31}$ = 6.3 $* 10^{- 2}$ g = 6.3 $* 10^{- 5}$ kg

Hence, 6.3 $* 10^{- 5}$ kg of $H_{2}$ will yield the same value of $\frac{P V}{T}$

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8 months ago

Class 11th Motion in a Plane

4.14 In a harbour, wind is blowing at the speed of 72 km/h and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat?

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Vishal Baghel

Contributor-Level 10

4.14

The velocity of the boat $V_{b}$ = 51 km/h, the velocity of the wind $V_{w}$ = 72 km/h

Flag is fluttering in N-E direction, so the $V_{w}$ direction is N-E.

When the ship begins to sail towards North, the flag will flutter along the direction of relative velocity of the wind w.r.t. the boat.

The angle between $V_{w}$ & - $V_{b}$ = 90 $° + 45 °$

tan $β$ = $\frac{V_{b} s i n (90 + 45)}{V_{w} + V_{b} c o s (90 + 45)}$ = $\frac{51 s i n ? (90 + 45)}{72 + 51 c o s ? (90 + 45)}$ = $\frac{36.062}{35.937}$

$β =$ tan-1( $\frac{36.062}{35.937}$ ) = 45.099 $°$

Angle with respect to East direction = 45.099 $°$ - 45 $°$ = 0.099 $°$

Hence the flag will flutter almost due East.

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8 months ago

Class 11th Motion in a Plane

4.13 A man can swim with a speed of 4.0 km/h in still water. How long does he take to cross a river 1.0 km wide if the river flows steadily at 3.0 km/h and he makes his strokes normal to the river current? How far down the river does he go when he reaches the other bank?

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Vishal Baghel

Contributor-Level 10

4.13 Swimming speed = 4 km/h, Width of the river = 1.0 km

Time required to cross the river = $\frac{W i d t h o f t h e r i v e r}{S w i m m i n g s p e e d}$ = $\frac{1}{4}$ h= 15 min

Speed of the river = 3 km/h

Distance covered by the river in 15 min = 3 $* (\frac{15}{60})$ $* 1000$ = 750 m

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8 months ago

Class 11th Motion in a Plane

4.12 Rain is falling vertically with a speed of 30 m s^-1. A woman rides a bicycle with a speed of 10 m s^-1 in the north to south direction. What is the direction in which she should hold her umbrella?

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Vishal Baghel

Contributor-Level 10

4.12 Let Vc = Velocity of the cyclist

Vr = Velocity of the rain

If V is the relative velocity, the woman must hold her umbrella in the direction of V.

V = Vr + (-Vc) = 30 + (-10) = 20 m/s

Tan $θ$ = $\frac{V c}{V r}$ = $\frac{10}{30}$ , $θ$ = 18 $°$

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8 months ago

Class 11th Kinetic Theory

13.2 Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0 °C). Show that it is 22.4 litres.

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Payal Gupta

Contributor-Level 10

13.2 The ideal gas equation relating to pressure (P), volume (V) and absolute temperature (T) is given by the relation

PV = nRT, where

R is the universal gas constant = 8.314 J /mol/K

N = number of moles = 1 (given)

T = standard temperature = 273 K

P = standard pressure = 1 atm = 1.013 $* 10^{5}$ N/ $m^{2}$

Therefore V = $\frac{n R T}{P}$ = $\frac{1 * 8.314 * 273}{1.013 * 10^{5}}$ = 0.02240 $m^{3}$ = 22.4 litres

Hence the molar volume of a gas at STP is 22.4 litres.

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8 months ago

Class 11th Motion in a Plane

4.11 A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi (b) the magnitude of average velocity? Are the two equal?

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Vishal Baghel

Contributor-Level 10

4.11

(a) Total distance travelled = 23 km, Total time taken = 28 min

Average speed = $\frac{T o t a l d i s t a n c e t r a v e l l e d}{T o t a l t i m e t a k e n}$ = $\frac{23}{28}$ km/min = 49.29 km/h

(b) Average velocity = $\frac{T o t a l d i s p l a c e m e n t}{t o t a l t i m e t a k e n}$ = $\frac{10}{28}$ km/min = 21.43 km/h

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8 months ago

Class 11th Motion in a Plane

4.10 On an open ground, a motorist follows a track that turns to his left by an angle of 60 $°$ after every 500 m. Starting from a given turn; specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.

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Vishal Baghel

Contributor-Level 10

4.10

Since the motorist taking the 60° turn after every 500m, It will form a perfect hexagon of side 500m after 6 turns.

Suppose the motorist starts from point A. From A, turns 60° to reach B (first turn), then to C and so on.

1. At third turn, The magnitude of displacement = DG + GA = 500m + 500m = 1000m Total path length = AB + BC + CD = 500m +500m +500m = 1500 m

2. At sixth turn, When the motorist takes the 6th turn, he is at starting point A. Therefore,

The magnitude of displacement = 0Total path length = AB + BC + CD + DE + EF + FA = 500m + 500m

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8 months ago

Class 11th Motion in a Plane

4.9 A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 4.21. If the round trip takes 10 min, what is the

(a) Net displacement

(b) Average velocity

(c) Average speed of the cyclist?

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Vishal Baghel

Contributor-Level 10

4.9

(a) Since the cyclist came back to his starting point O, so the displacement is Zero

(b) Average velocity = $\frac{n e t d i s p l a c e m e n t}{t i m e t a k e n} = \frac{0}{t i m e t a k e n}$ = 0

(c) Average speed = $\frac{d i s t a n c e t r a v e l l e d}{t i m e t a k e n}$

Distance travelled = OP + PQ + QO

= 1 + (1/4) $* 2 * (\frac{22}{7}) * 1 + 1 =$ 3.57 km

Average speed = 3.57 / 10 km/min = 21.43 km/h

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8 months ago

Class 11th Motion in a Plane

4.8 Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. 4.20. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of path skate?

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Vishal Baghel

Contributor-Level 10

4.8 The radius of the ground = 200 m, the distance PQ = diameter of the ground = 400 m

The displacement all 3 girls = 400 m

This magnitude is equal to the path skated by girl B

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