Question

please answer 2,3,4,5,6,7,8 thankyou

2. If the ambulance siren has a frequency of 800Hz, then to a person hearing the approaching siren, the frequency would be a) Exactly 680Hz b) Exactly 800Hz c) Lower than 800Hz d) Higher than 800Hz 3. To find the focal length of a concave facial mirror, a person would hold the mirror a) Far away and slowly bring it closer until the image becomes real and upright b) Far away and slowly bring it closer until the image becomes virtual and inverted c) Close initially, and then move it away slowly until the image becomes real and inverted d) Close initially, and then move it away slowly until the image becomes virtual and inverted 4. The object in the diagram is 2.50cm from the objective lens, which has a focal length of 2.00cm. The eyepiece is placed 12.0cm from the objective lens and has a focal length of 2.3cm. Taking right to be positive, the distance between the final image and the eyepiece is a) 15.3cm b) 10.0cm c) -10.0cm d) -15.3cm 5. An object is travelling at a constant speed of 0.900c. In comparison with its rest state, to a stationary observer the object will appear a) Less massive and longer b) Less massive and shorter c) More massive and longer d) More massive and shorter 6. A light bulb receives power from an external source at a rate of 40.0W. If the light bulb is 58.0% efficient, how much energy is produced as light in 10.0 minutes? a) 1.15x109) b) 1.23x109) c) 1.39x109) d) 1.46x109)

Answer 1

2) we have a formula from doppler effect,

$f'=\frac{(v+v_0)}{(v-v_s)}f$

where
f'
is observed frequency, f is actual frequency, v is sound velocitty, $v_s$ is source velocity and $v_0$ is observer velocity. Here in that question observer is at rest and source coming towards observer, so $v_0$ is zeo and $v_s$ is positive. From the equation we can clearly say that observed frequency will be higher than siren frequency 800Hz.

So option d is answer.

3) Option C is the right answer. As the image by the concave mirror will be real and inverted.

4)

5) The relativistic mass formula,

heyepiece, Fo Fe ohject to Fe -final mage thi objective Cerise dili de = the distance bo final emage & eyepiece. distance be object & objective dense focal length of objective lens. fea focal length of exepiece. do = the Fo we have do=2.50 cm fo = 2.00 cm. fe - 2.3 cm. ditdó 12 am Wehave, di = /fo-do = 0.1 cm di =10cm, = -0.065217 391 emt. then, dó = (12-10) cm = 2cm. I fe- lao di = option, (d) is answer di' - 15.33 cm.

$m_{rel}=\frac{m}{\sqrt(1-\frac{v^2}{c^2})}$

where m is actual mass, v is velocity of object.

And for the relativistic length,

$L_{rel}=L_0{\sqrt(1-\frac{v^2}{c^2})}$

where $L_0$ is actual length.

For v = 0.9c

${\sqrt(1-\frac{v^2}{c^2})} = 0.436$

SO it is clear that object will appear massive and length will contacted that is shorter.

So option (d) is correct.

7)

Re А 7. Rt Evi 3 R3 tarih In Series R & R2 are in Comection & Rz & Ry arem parallel connection. so, for the sessie's connection of resistance R & R2 voltage will be added. & form parallel connection of Rz & Rz. voltage will be same at point Voltage between points A&B. is V3 = 4.6 V=V, + V₂ + V3. (and be barof

6) The answer of question 6 is (c).

of Wout x 100'/. Win 1889220.0 sucheus by Efficiency of a bulb, no Wout energy output hale . energy input rate. We have, win= Win=yow. & n = 58% : Wout = (58 x 40 )w. = 23.2 7/sec . In 10min, energy produced by the bulbis, fout = (23.26X10X60) I. (Amd. = 232@w = 23-20 J/sec. 100 J PH020 to 4 1.39 X10

8) Option (a) is correct.

$4\Omega$ is in parrallel connection to $2\Omega$ so effective resistance is lower than $2\Omega$ .

As we remove $4\Omega$ resistance The effective resistance of the circuit will increase.

Then voltage across $1\Omega$ will increase also V will increase.