UNITS AND MEASUREMENTS Quiz-12
Dear Readers,
As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background.
Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced.
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Q1. In a new system of units, unit f mass is 10 kg, unit of length is 1 km and unit of time is 1 min. The value of 1 joule in this new hypothertical system is
Solution
We know that the dimensional formula of energy is [ML2 T-2] n_2=1[[1kg/10kg]-1 [1m/1km] [1s/(1 min)]2 =1/10×1/106 ×1/(60)-2 =3600/107 =3.6×10-4
We know that the dimensional formula of energy is [ML2 T-2] n_2=1[[1kg/10kg]-1 [1m/1km] [1s/(1 min)]2 =1/10×1/106 ×1/(60)-2 =3600/107 =3.6×10-4
Q2.In an experiment, the following observation’s were recorded: L=2.820 m, M=3.00 kg,l=0.087 cm, diameter D=0.041 cm. Taking g=9.81 m/s2 using the formula,Y=4MgL/(πD2 l), the maximum permissible error in Y is
Solution
Y=4MgL/(πD2 I) so maximum permissible error in Y =∆Y/Y×100=(∆M/M+∆g/g+∆L/L+2∆D/D+∆l/l)×100 =(1/300+1/981+1/2820+2×1/41+1/87)×100 =0.065×100=6.5%
Y=4MgL/(πD2 I) so maximum permissible error in Y =∆Y/Y×100=(∆M/M+∆g/g+∆L/L+2∆D/D+∆l/l)×100 =(1/300+1/981+1/2820+2×1/41+1/87)×100 =0.065×100=6.5%
Q3. Which relation is wrong
Solution
1 newton = 10-5 dyne
1 newton = 10-5 dyne
Q4. The resistance R=V/i where V=100±5 volts and i=10±0.2 amperes. What is the total error in R
Solution
∴ (∆R/R×100)max=∆V/V×100+∆I/I×100 =5/100×100+0.2/10×100=(5+2)%=7%
∴ (∆R/R×100)max=∆V/V×100+∆I/I×100 =5/100×100+0.2/10×100=(5+2)%=7%
Q1. The quantity X=(ε0 LV)/t:ε0 is the permittivity of free space, L is length, V is potential difference and t is time. The dimensions of X are same as that of
Solution
[ε0 L]=[C]∴X=(ε0 LV)/t=(C×V)/t=Q/t=current
[ε0 L]=[C]∴X=(ε0 LV)/t=(C×V)/t=Q/t=current
Q6. Ampere-hour is a unit of
Solution
Charge =current ×time
Charge =current ×time
Q7. 1 Wb/m2 is equal to
Solution
If a charge of 1 C moving with a velocity of 1 ms-1 perpendicular to a uniform magnetic field experiences a force of 1 N, then the magnitude of the field is 1 T. The SI unit of magnetic field is Wb m-2. Thus, 1 T=1 NA-1 m-1=1 Wbm-2 In CGS systems 1 tesla=10-4 gauss=1 Wbm-2
If a charge of 1 C moving with a velocity of 1 ms-1 perpendicular to a uniform magnetic field experiences a force of 1 N, then the magnitude of the field is 1 T. The SI unit of magnetic field is Wb m-2. Thus, 1 T=1 NA-1 m-1=1 Wbm-2 In CGS systems 1 tesla=10-4 gauss=1 Wbm-2
Q8. Unit of magnetic moment is
Solution
M=Pole strength × length =amp-metre ×metre =amp-metre2
M=Pole strength × length =amp-metre ×metre =amp-metre2
Q9. Surface tension has the same dimensions as that of
Solution
Surface Tension = Force/Length
=[MLT-2 ]/([L])=[ML0 T-2] Spring constant = Force/Length =[MLT-2 ]/([L])=[ML0 T-2]
Surface Tension = Force/Length
=[MLT-2 ]/([L])=[ML0 T-2] Spring constant = Force/Length =[MLT-2 ]/([L])=[ML0 T-2]
Q10. The dimensions of physical quantity X in the equation
Force =X/(Density ) is given by
Solution
[X]=[F]×[ρ]=[MLT-2 ]×[(M/L3)]=[M2 L-2 T-2]
[X]=[F]×[ρ]=[MLT-2 ]×[(M/L3)]=[M2 L-2 T-2]
