Q1.Medical termination of pregnancy is considered safe up to _________ of pregnancy.
Answer : Option AExplaination / Solution:
Medical termination of pregnancy is considered safe up to twelve weeks or three months. At the end of three months all organs of foetus are formed. It is fetal to mother in case of MTP after three months.
Q2.The resistance of a metallic conductor increases due to
Answer : Option CExplaination / Solution: Resistance of a metal of length l , area A is R=mlne2τA where n is the carrier density, m the mass of the electron , e the charge of the electron and τ, the average time between two collisions. at constant temperature, n and τ remain constant. mne2τ is constant at constant temperature and is called ∴R=ρlA The resistance increases if the length increases or the area decreases.
Q3.A beam of alpha particles is incident on a target of lead. A particular alpha particle comes in “head-on” to a particular lead nucleus and stops 6.5 ×10−14m away from the center of the nucleus. (This point is well outside the nucleus.) Assume that the lead nucleus, which has 82 protons, remains at rest. The mass of the alpha particle is 6.64 ×10−27kg . Calculate the electrostatic potential energy at the instant that the alpha particle stops. Express your result in MeV.
Answer : Option BExplaination / Solution: No Explaination.
Q6.A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is similarly charged to a potential difference 2V. The charging battery is then disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is
Answer : Option DExplaination / Solution: The charges Q1 and Q2 on the two capacitors Q1=CV;Q2=(2C)(2V)=4CV.The capacitors are connected in parallel in such a way that the positive plate of one is connected to the negative plate of the other. The common potentialV=Q2−Q1C+2C=4CV−CV3C=V.The final energy Uf=12CV2+12(2C)V2=32CV2
Q9.A thin uniform rod of length 2l and mass M is acted upon a constant torque. The angular velocity changes from zero to ωω in time t. The value of torque is
Answer : Option BExplaination / Solution: As Torque(τ) is equal to product of Moment of Inertia (I) and Angular acceleration (α)