Q4
Standard deductions under Section 24 in income from house property income is
సెక్షన్ 24 ప్రకారం గృహస్తి ఆదాయంలో ప్రామాణిక తగ్గింపు మొతం
Ops: A. Oa) 30% of NAV
ఎ) నికర వార్షిక విలువలో 30%
Ob) 25% of NAV
బి) నికర వార్ధిక విలువలో 25%
Oc)20% of NAV
(సి) నికర వార్షిక విలువలో 20%
D. Od) 10% of NAV
డి) 10% నికర వార్షిక విలువలో
Answers
Answer:
a.) 30% of NAV
Section 24 of the Income Tax Act lets homeowners claim a deduction of up to Rs. 2 lakhs (Rs. 1,50,000 if you are filing returns for last financial year) on their home loan interest if the owner or his family reside in the house property. The entire interest is waived off as a deduction when the house is on rent.
hope it helps u
\begin{gathered}\begin{gathered} \\ \Large{\bf{\pink{\underline{Answer \::}}}} \\ \end{gathered} \end{gathered}
Answer:
We know that kinetic energy is given by,
K.E = 1/2 × m × v²
where m is the mass and v is the velocity of the body.
But we know,
,
\tt v=\pm \omega\sqrt{A^2-x^2}vv=±ω
A
2
−x
2
v
Hence,
\tt K.E=\dfrac{1}{2} \times m\times \omega^{2}\times (A^2-x^2)K.E=K.E=
2
1
×m×ω
2
×(A
2
−x
2
)K.E=
Also,
ω² = k/m
k = ω²m
Substituting this we get,
\boxed{ \red{\tt K.E=\dfrac{1}{2} \times k\times (A^2-x^2)} }
K.E=
2
1
×k×(A
2
−x
2
)
Now deducing the equation for potential energy. Here work done by the particle is stored as potential energy.
dW = -f dx
dW = -(-kx) dx
dW = kx dx
Integrating this we from position 0 to x get,
get,
\displaystyle \sf \int\limits {dW}=\int\limits^x_0 k{x} \,∫dW=
0
∫
x
kx
W = k x²/2
\boxed{ \red{\tt P.E=\dfrac{1}{2}\times k\times x^2 } } < /p > < p >
P.E=
2
1
×k×x
2
</p><p>
Now total energy of the particle is given by,
T.E = K.E + P.E
\tt { \red{T.E=\dfrac{1}{2}\:k\times (A^2-x^2)+ \dfrac{1}{2}\: k\times x^2 }}T.E=
2
1
k×(A
2
−x
2
)+
2
1
k×x
2
\tt{ \red{T.E=\dfrac{1}{2}\:k(A^2-x^2+x^2)}}T.E=
2
1
k(A
2
−x
2
+x
2
)
\boxed{ \red{\tt T.E=\dfrac{1}{2}\:k\:A^2} }
T.E=
2
1
kA
2
Here k and A are constants for a given SHM motion, therefore the total energy is conserved.
Factors on which total energy depends:
From the above expression,
\tt \red{T.E=\dfrac{1}{2}\:k\:A^2 }T.E=
2
1
kA
2
\tt \red{T.E=\dfrac{1}{2}\: m\: \omega^{2} \:A^2}T.E=
2
1
mω
2
A
2
Hence total energy depends upon the mass, angular velocity and amplitude of the particle.