Human activities are prohibited to save forest and wildlife in * National Parks Biosphere Reserves Wildlife sanctuaries All of the above

Explanation:

Classes :-

5-7 , 7-9 , 9-11 , 11-13 , 13-15 , 15-17

Frequency :-

55, 55, 70, 150, 86, 84

Cumulative frequencies :-

55, 110, 180, 330, 416 , 500

Sum of all frequencies (N)

= 55+55+70+150+86+84

= 500

Therefore, N = 500

N/2 = 500/2 = 250

250 lies in the cumulative frequency 330

Frequency of Median class = 150

Class interval of Median class = 11-13

Lower boundary of the median class (l) = 11

[tex]\begin{gathered}\boxed{\begin{array}{c|c|c} \bf Classes & \bf Frequency & \bf Cumulative\; Frequency \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 5 - 7 & 55& 55 \\ \\ \sf 7 - 9& 55 & 110\\ \\ \sf \: 9 - 11& \sf 70 & \sf 180\\ \\ 11 - 13 & 150 & 330 \\ \\ \sf 13 - 15 & \sf 86 & 416\\ \\ \sf \: 15 - 17& \sf \: 84 & \sf 500 \end{array}} \\ \end{gathered}[/tex]

To find the median class:

• ➡ First add the total number of frequencies.

[tex]\rm \implies \: 55 +55+ 70 + 150 + 86+84 [/tex]

[tex]\bf\implies \: 500[/tex]

Here , we know that

• ➡ f = 500

Now ,

[tex]\rm \implies \: \dfrac{(Total \: number \: of \: frequency)}{2} [/tex]

• Put the given values in this and solve

[tex]\rm \implies \dfrac{500}{2} = 250[/tex]

Henceforth,

• ➡ The value 250 lies in the class interval 11-13 from the cummulative frequency table.

• ➡ As 330 is just greater than 250,

[tex]\rm \longrightarrow \: \therefore \: Median \: class = 11 - 13[/tex]

[tex] \rm \therefore \: The \: lower \: limit \: of \: the \: modal \: class \: is \: 11.[/tex]

Answer:

a. penna

Explanation:

the other river are in india

(a) Brahmaputra

The river Brahmaputra is located in the eastern mountain ranges of India while the rest of the mentioned rivers flow in the western and southern plateau regions.

At an elevation of 5300 metres, the river begins in the Himalayan Kailash mountains. It enters India through Arunachal Pradesh and then flows through Assam and Bangladesh before joining the Bay of Bengal.

Answer:

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Answer:

Principle: 

An electric motor is a rotating device that converts electrical energy to mechanical energy working.

Working: 

Current in the coil ABCD enters from the source battery through conducting brush X and flow back to the battery through brush Y. Notice that the current in the arm AB of the coil flows from A to B. In arm CD it flows from C to D that is opposite to the direction of current through arm AB on applying Fleming’s left hand rule for the direction of force on a current-carrying conductor in a magnetic field. We find that the force acting on arm AB pushes it downwards while the force acting on arm CD pushes it upwards.

Thus the coil and the Axle O, mounted free to turn about an axis, rotate anti-clockwise at half rotation. Q makes contact with the brush X and P with brush Y. Therefore the current in the coil gets reversed and flows along the path DCBA. The reversal of current also reverses the direction of force acting on the two arms AB and CD. Thus the arm AB of the coil that was earlier pushed down, is now pushed up and the arm CD previously pushed up is pushed down. There is a continuous rotation of the coil and to the axle. 

Split rings in electric motors acts as a commutator.

Explanation:

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Answer:

a current carrying conductor when placed at right angle to a magnetic field experience of force due to which we get motion the direction of force is given by felmings right hand rule

Explanation: