Prasar Bharati Combined Recruitment for the post of Engineering Assistant & Technician Examination – 2013

Answer key with complete solutions of test form no. 555QJ4

 

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Technical knowledge section

 

101)  Ans (C)

Clouds have low thermal conductivity.It will not allow heat to escape easily.

 

102)  Ans (B)

P_x = P_Total ( n_x / n_Total )where
P_x = partial pressure of gas x
P_Total = total pressure of all gases
n_x = number of moles of gas x
n_Total = number of moles of all gases
P_hydrogen = P_Total ( n_hydrogen / n_Total )
P_helium = P_Total ( n_helium / n_Total )
P_hydrogen / P_helium  = n_hydrogen  /  n_helium
                                                = 2 / 4 = 0.5

103)  Ans (C)

Consider an electron moving from north to south on the plane of a paper. Electric field direction is downwards, which means positive charge is above plane of paper and negative is below it. Since electron always moves towards positive, its direction will be upward.

104)   Ans (B)

 E=q/(4πεr²)

Ratio for r,2r,3r will be  1: (1/4) : (1/9) which is 36 : 9 : 4

105)   Ans (C)

Topic - optics

Distance between slits , y = nλd/D

n – maxima number; λ – wavelength ; d – distance of slits from screen; D - distance between slits

106)  Ans (C)

Frequency cannot change with medium

107)  Ans(B)

KE = (3/2)KT

KE depends only on temperature

108) Ans(B)

109) Ans(D)

110) Ans(C)

X_c = 1/(2πfc)

5 = 1/(2π*50*C) ; find C and Solve for X_c  at 200Hz

111) Ans(not in choices)

Topic – Doppler effect

For source moving towards observer, f’ = v/v-v_s   * f

For source moving away from observer, f’ = v/v+v_s   * f

112) Ans(D)

Topic – torsion pendulum

Couple per unit twist C = ½ πnr⁴ / l

Where r –radius of rod ; l- length of rod ; n- rigidity modulus

113) Ans (A)

114) Ans (D)

Mean life = 1/ (decay constant)

115) Ans(C)

F_m = 5KHz

BW = 2 *5 =10 KHz

Number of stations = 100/10 =10

116) Ans(C)

β = α / 1-α

117) Ans(D)

The signals are first passed through filters which only allow through frequencies up to 15 kHz. The L and R signals are then added to produce a sum signal and subtracted one from the other to produce a difference signal. The sum is essentially a monophonic signal which is what we would send for playing through a single loudspeaker. The difference signal is used to DSBSC modulate a 38 kHz sinewave.The DSBSC output is added to the sum (mono) signal and the combination is sent on the transmitter's FM modulator. A monophonic receiver can now ignore the stereo information simply by using a filter after its FM demodulator to block everything above 15 kHz. A stereo receiver has to have an additional circuit after the FM demodulator which can detect and demodulate the DSBSC wave.

118) Ans(B)

Local oscillator frequency F_o  = f + f_IF

119)  Ans(C)

Magnification=Focal length of the Objective/Focal length of the Eyepiece

120) Ans (C)

Dielectric constant = ε / ε₀

121) Ans(B)

Area of B-H loop gives energy dissipated on reversing magnetic field.

In 50Hz cycles magnetisation changes @ 50 times/sec. In 1 hour it changes 60*60*50 times

Energy dissipated = 10*60*60*50

122) Ans(B)

Efficiency = output power/input power = I²R/VI =  5² * 10 / 200 * 5

123) Ans(C)

Topic - Thomson effect

Heat released/absorbed , H = σ Q Δθ

Q – charge, σ – Thomson coefficient, Δθ – temperature difference

124) Ans(A)

125) Ans(D)

Stoneman transmission bridge is used in telephone communication (at central offices) to separate voice path from signal path (used for ring tone, dial tone, busy etc) and battery.

126) Ans(C)

Grade of service of a telephone system = number of blocked calls  / total number of calls

127) Ans(B)

128) Ans(B)

AM range 535KHz – 1635KHz

129) Ans(D)

130) Ans(D)

 %Modulation = V_m / V_c   * 100  

131) Ans(B)

132) Ans(A)

Mixer gives sum and difference frequencies at the output.

133) Ans(A)

Microwave range

134) Ans(D)

135) Ans(A)

Ratio detector provides inbuilt  AGC mechanism

136) Ans(D)

137) Ans(C)

138) Ans(C)

139) Ans(D)

140) Ans(C)

f_s = 2f =200 Hz

141) Ans(B)

142) Ans(C)

Preemphasis – boosting higher frequencies

143) Ans(C)

144) Ans(D)

All others are light emitters

145) Ans(B)

146) Ans(C)

R=G=0 for lossless line

147) Ans(B)

Compressor –expandor.

On compression dynamic range is limited. Only few quantization levels are needed. step size  can be reduced and thus quantization noise .

148) Ans(B)

149) Ans(C)

Video transmission – AM

Audio transmission – FM

150) Ans(C)

Higher frequency => narrow beam => accurate focussing a satellite in the crowded satellite space