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Page 2 As we know, for Bipolar junction transistor The length of each section of transistor is in order as shown above. i.e., L C > L E > L B and doping profile is Emitter > Collector > Base Whereas, for transistor action Baseemitter junction is forward biased and Basecollector junction is reversed biased. Hence from this we can conclude that, for transistor action the base region must be very thin and lightly doped. Q 2. A spherical conductor of radius 10 cm has a charge of 3.2 × 10 –7 C distributed uniformly. What is the magnitude of electric field at a point 15 cm from the centre of the sphere? ?? ???? ?? ?? ?? 9 2 2 0 1 9 10 Nm / C 4 Option A 1.28 × 10 6 N/C Option B 1.28 × 10 7 N/C Option C 1.28 × 10 4 N/C Option D 1.28 × 10 5 N/C Correct Option D Solution: For a conducting sphere the electric field outside will be given as Q 3. Assume that light of wavelength 600 nm is coming from a star. The limit of resolution of telescope whose objective has a diameter of 2 m is _______. Option A 7.32 × 10 –7 rad Option B 6.00 × 10 –7 rad Option C 3.66 × 10 –7 rad Option D 1.83 × 10 –7 rad Correct Option C Solution: As we know, Page 3 As we know, for Bipolar junction transistor The length of each section of transistor is in order as shown above. i.e., L C > L E > L B and doping profile is Emitter > Collector > Base Whereas, for transistor action Baseemitter junction is forward biased and Basecollector junction is reversed biased. Hence from this we can conclude that, for transistor action the base region must be very thin and lightly doped. Q 2. A spherical conductor of radius 10 cm has a charge of 3.2 × 10 –7 C distributed uniformly. What is the magnitude of electric field at a point 15 cm from the centre of the sphere? ?? ???? ?? ?? ?? 9 2 2 0 1 9 10 Nm / C 4 Option A 1.28 × 10 6 N/C Option B 1.28 × 10 7 N/C Option C 1.28 × 10 4 N/C Option D 1.28 × 10 5 N/C Correct Option D Solution: For a conducting sphere the electric field outside will be given as Q 3. Assume that light of wavelength 600 nm is coming from a star. The limit of resolution of telescope whose objective has a diameter of 2 m is _______. Option A 7.32 × 10 –7 rad Option B 6.00 × 10 –7 rad Option C 3.66 × 10 –7 rad Option D 1.83 × 10 –7 rad Correct Option C Solution: As we know, The limit of resolving power is given as Q 4. Dimensions of stress are: Option A [ML 0 T –2 ] Option B [ML –1 T –2 ] Option C [MLT –2 ] Option D [ML 2 T –2 ] Correct Option B Solution: ?? ?? ?? ?? 2 2 1 2 Force Stress= Area MLT = L = ML T Q 5. A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale. The pitch of the screw gauge is _______. Option A 0.5 mm Option B 1.0 mm Option C 0.01 mm Option D 0.25 mm Correct Option A Solution: Least count of screw gauge Pitch = Number of divisions on circular scale Pitch 0.01 mm = 50 Pitch = 0.5 mm Q 6. Two bodies of mass 4 kg and 6 kg are tied to the ends of a massless string. The string passes over a pulley which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity (g) is _______. Page 4 As we know, for Bipolar junction transistor The length of each section of transistor is in order as shown above. i.e., L C > L E > L B and doping profile is Emitter > Collector > Base Whereas, for transistor action Baseemitter junction is forward biased and Basecollector junction is reversed biased. Hence from this we can conclude that, for transistor action the base region must be very thin and lightly doped. Q 2. A spherical conductor of radius 10 cm has a charge of 3.2 × 10 –7 C distributed uniformly. What is the magnitude of electric field at a point 15 cm from the centre of the sphere? ?? ???? ?? ?? ?? 9 2 2 0 1 9 10 Nm / C 4 Option A 1.28 × 10 6 N/C Option B 1.28 × 10 7 N/C Option C 1.28 × 10 4 N/C Option D 1.28 × 10 5 N/C Correct Option D Solution: For a conducting sphere the electric field outside will be given as Q 3. Assume that light of wavelength 600 nm is coming from a star. The limit of resolution of telescope whose objective has a diameter of 2 m is _______. Option A 7.32 × 10 –7 rad Option B 6.00 × 10 –7 rad Option C 3.66 × 10 –7 rad Option D 1.83 × 10 –7 rad Correct Option C Solution: As we know, The limit of resolving power is given as Q 4. Dimensions of stress are: Option A [ML 0 T –2 ] Option B [ML –1 T –2 ] Option C [MLT –2 ] Option D [ML 2 T –2 ] Correct Option B Solution: ?? ?? ?? ?? 2 2 1 2 Force Stress= Area MLT = L = ML T Q 5. A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale. The pitch of the screw gauge is _______. Option A 0.5 mm Option B 1.0 mm Option C 0.01 mm Option D 0.25 mm Correct Option A Solution: Least count of screw gauge Pitch = Number of divisions on circular scale Pitch 0.01 mm = 50 Pitch = 0.5 mm Q 6. Two bodies of mass 4 kg and 6 kg are tied to the ends of a massless string. The string passes over a pulley which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity (g) is _______. Option A g/5 Option B g/10 Option C g Option D g/2 Correct Option A Solution: For the given case, the acceleration of given system will be expressed as Q 7. An electron is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength of the electron is 1.227 × 10 –2 nm, the potential difference is : Option A 103 V Option B 104 V Option C 10 V Option D 102 V Correct Option B Solution: According to de Broglie equation, the wavelength of electron will be given as ? o 10 2 11 4 12.27 ? = A V 12.27×10 V = = 10 1.227×10 V = 10 volts Page 5 As we know, for Bipolar junction transistor The length of each section of transistor is in order as shown above. i.e., L C > L E > L B and doping profile is Emitter > Collector > Base Whereas, for transistor action Baseemitter junction is forward biased and Basecollector junction is reversed biased. Hence from this we can conclude that, for transistor action the base region must be very thin and lightly doped. Q 2. A spherical conductor of radius 10 cm has a charge of 3.2 × 10 –7 C distributed uniformly. What is the magnitude of electric field at a point 15 cm from the centre of the sphere? ?? ???? ?? ?? ?? 9 2 2 0 1 9 10 Nm / C 4 Option A 1.28 × 10 6 N/C Option B 1.28 × 10 7 N/C Option C 1.28 × 10 4 N/C Option D 1.28 × 10 5 N/C Correct Option D Solution: For a conducting sphere the electric field outside will be given as Q 3. Assume that light of wavelength 600 nm is coming from a star. The limit of resolution of telescope whose objective has a diameter of 2 m is _______. Option A 7.32 × 10 –7 rad Option B 6.00 × 10 –7 rad Option C 3.66 × 10 –7 rad Option D 1.83 × 10 –7 rad Correct Option C Solution: As we know, The limit of resolving power is given as Q 4. Dimensions of stress are: Option A [ML 0 T –2 ] Option B [ML –1 T –2 ] Option C [MLT –2 ] Option D [ML 2 T –2 ] Correct Option B Solution: ?? ?? ?? ?? 2 2 1 2 Force Stress= Area MLT = L = ML T Q 5. A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale. The pitch of the screw gauge is _______. Option A 0.5 mm Option B 1.0 mm Option C 0.01 mm Option D 0.25 mm Correct Option A Solution: Least count of screw gauge Pitch = Number of divisions on circular scale Pitch 0.01 mm = 50 Pitch = 0.5 mm Q 6. Two bodies of mass 4 kg and 6 kg are tied to the ends of a massless string. The string passes over a pulley which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity (g) is _______. Option A g/5 Option B g/10 Option C g Option D g/2 Correct Option A Solution: For the given case, the acceleration of given system will be expressed as Q 7. An electron is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength of the electron is 1.227 × 10 –2 nm, the potential difference is : Option A 103 V Option B 104 V Option C 10 V Option D 102 V Correct Option B Solution: According to de Broglie equation, the wavelength of electron will be given as ? o 10 2 11 4 12.27 ? = A V 12.27×10 V = = 10 1.227×10 V = 10 volts Q 8. In a certain region of space with volume 0.2 m 3 , the electric potential is found to be 5 V throughout. The magnitude of electric field in this region is : Option A 1 N/C Option B 5 N/C Option C zero Option D 0.5 N/C Correct Option C Solution: For the given case, since electric potential is found constant throughout the region. Hence by using the equation of electric field. i.e., dV E =  = 0 dr We can conclude that the magnitude of electric field in the given region will be zero Q 9. A cylinder contains hydrogen gas at pressure of 249 kPa and temperature 27°C. Its density is : (R = 8.3 J mol –1 K –1 ) Option A 0.1 kg/m 3 Option B 0.02 kg/m 3 Option C 0.5 kg/m 3 Option D 0.2 kg/m 3 Correct Option D Solution: Q 10. The mean free path for a gas, with molecular diameter d and number density n can be expressed as : Option A Option B Option C Option DRead More
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NEET 2019 Past Year Paper with Detailed Solutions Doc  69 pages 
NEET 2018 Past Year Paper with Detailed Solutions Doc  70 pages 
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NEET 2019 Past Year Paper with Detailed Solutions Doc  69 pages 
NEET 2018 Past Year Paper with Detailed Solutions Doc  70 pages 
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