Answer
Learner's license expires after six months of the date of issue
Explanation:
The learner's license issued to make a person eligible to learn driving in a given time period. In India the learning license is valid for six months after the issuing date but the person have to apply for the driving license within the three months of the issued date. You can drive the vehicles listed in the learner's license document even after the three months of the issuing date. The process of obtaining a driving license have certain process
- Go to the online web portal of your state driving license authority apply for the driving license and print the application form page from there.
- Attach this application form and RED CROSS certificate with the file given by the R.T.O office and submit this to them.
- The authority will give you date, time and site of the driving test where you have to got there with your vehicle for the test.
- If you passes the driving test your file will pass and you can take your driving license after some days from the R.T.O office.
Answer:
The given blanks can be filled as given below
Voltmeter must be connected in parallel
Explanation:
A voltmeter is connected in parallel to measure the voltage drop across a resistor this is because in parallel connection, current is divided in each parallel branch and voltage remains same in parallel connections.
Therefore, in order to measure the same voltage across the voltmeter as that of the voltage drop across resistor, voltmeter must be connected in parallel.
Given that,
Mass of the object 1, m = 107.01 grams
To find,
Force on the object.
Solution,
The force acting on the object is gravitational force. The force is given by the formula as follow :
F = mg
g is acceleration due to gravity
F = 0.10701 kg × 9.8 m/s²
F = 1.048 N
So, the force acting on object 1 is 1.048 N.
I think the answer to this question is b but I’m not sure
Answer:
Given that
LHS of above given equation have dimension ![[M^oL^{1}T^{-1}]](https://tex.z-dn.net/?f=%5BM%5EoL%5E%7B1%7DT%5E%7B-1%7D%5D)
.
Now find the dimension of RHS
Dimension of P = ![[ML^{-1}T^{-2}]](https://tex.z-dn.net/?f=%5BML%5E%7B-1%7DT%5E%7B-2%7D%5D)
.
Dimension of d= ![[M^{0}L^{1}T^{0}]](https://tex.z-dn.net/?f=%5BM%5E%7B0%7DL%5E%7B1%7DT%5E%7B0%7D%5D)
.
Dimension of μ = ![[ML^{-1}T^{-1}]](https://tex.z-dn.net/?f=%5BML%5E%7B-1%7DT%5E%7B-1%7D%5D)
.
Dimension of L= .
So
![\dfrac{\Delta Pd^2}{32\mu L}=\dfrac{[ML^{-1}T^{-2}].[M^{0}L^{1}T^{0}]^2}{[ML^{-1}T^{-1}].[M^{0}L^{1}T^{0}]}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5CDelta%20Pd%5E2%7D%7B32%5Cmu%20L%7D%3D%5Cdfrac%7B%5BML%5E%7B-1%7DT%5E%7B-2%7D%5D.%5BM%5E%7B0%7DL%5E%7B1%7DT%5E%7B0%7D%5D%5E2%7D%7B%5BML%5E%7B-1%7DT%5E%7B-1%7D%5D.%5BM%5E%7B0%7DL%5E%7B1%7DT%5E%7B0%7D%5D%7D)
![\dfrac{\Delta Pd^2}{32\mu L}=[M^0L^{1}T^{-1}]](https://tex.z-dn.net/?f=%5Cdfrac%7B%5CDelta%20Pd%5E2%7D%7B32%5Cmu%20L%7D%3D%5BM%5E0L%5E%7B1%7DT%5E%7B-1%7D%5D)
It means that both sides have same dimensions.
