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P = constant

m (dv/dt) = P
∫? vdv = P/m ∫? dt
v²/2 = Pt/m ⇒ v = (2Pt/m)¹/²
dx/dt = √ (2P/m) t¹/²
∫? dx = √ (2P/m) ∫? t¹/² dt
x = √ (2P/m) (t³/² / (3/2) = √ (2P/m) × (2/3) × t³/²
= (2/3) × 27 = 18

V_rms = √ (3RT/M)
V_N? = V_H?
√ (3RT_N? / M_N? ) = √ (3RT_H? / M_H? )
573/28 = T_H? /2
⇒ T_H? = 40.928

δ_min = (µ - 1)A
= (1.5 - 1)1
= 0.5
δ_min = 5/10
N=5

$=\frac{1}{56}\times 277.85=496$

$\mathrm{}\pm 1=\left|\begin{array}{lll}1& 1& \lambda \\ 1& 1& 3\\ 2& 1& 1\end{array}\right|\Rightarrow =-\lambda +3=\pm 1$ or $\Rightarrow \lambda =2$

For $\lambda =4$ $$

$\lambda =4$

$\mathrm{c}\mathrm{o}\mathrm{s}?\theta =\frac{2+1+4}{\sqrt[]{6}\sqrt[]{18}}=\frac{7}{6\sqrt[]{3}}$

E? = φ + K?
E? = φ + K?
E? - E? = K? - K?
Now V? /V? = 2
K? /K? = 4; K? = 4K?
Now from equation (2)
⇒ 4 - 2.5 = 4K? - K?
1.5 = 3K?
K? = 0.5eV
Now putting this
Value in equation (2)
2.5 = φ + 0.5eV
φ = 2eV

Yes, the CHTS Institute of Hotel Management, Catering and Tourism accept Class 12 marks for UG courses. Candidates seeking admission to all programme can enrol for admission with Class 12 marks. Candidates must complete their Class 12 to enrol to the UG courses. The institute offers UG, PG, Diploma and PG Diploma courses.

L_i = L_f
mvL = Iω
mvL = (ML³/3 + mL²)ω

Before collision
After collision
0.1 × 80 × 1 = (0.9 × 1²)/3 + 0.1 × 1²)ω
8 = (3/10 + 1/10)ω 8 = (4/10)ω ω = 20 rad/sec

The core BA psychology subjects that you'll study in the programme are

• Introduction to Psychology 
• Statistical Methods for Psychological Research
• Biopsychology
• Psychology of Individual Differences
• Development of Psychological Thought
• Social Psychology
• Understanding Psychological Disorders 
• Developmental Psychology 
• Organizational Behavior 
• Organizational Behavior

Let us assume the potential at A = V_A = 0.
Now at junction C, according to KCL
i? + i? = i?
1 A + i? = 2 A

i? = 2 A
Now analyse potential along ACDB
V_A + 1 + i? (2) - 2 = V_B
0 + 1 + 2 (1) - 2 = V_B
V_B = 3 - 2
V_B = 1amp.

The tuition fee structure followed at MPSTME BTech courses is a per annum model, according to the fee data. The total tuition fee for BTech courses at MPSTME varies between BTech specialisations. In 2025, the total tuition fees at MPSTME can range between INR 14 Lacs - INR 25.1 lakh. All candidates taking admissions intoMPSTME BTech courses are advised to visit the official website to confirm the fee amount.

Admissions at the CHTS Institute of Hotel Management, Catering and Tourism are currently open. Candidates are selected for courses based on students' merit. CHTS Institute of Hotel Management, Catering and Tourism offers BVoc, MBA, etc. 

Figure of Merit = C = i/θ
= C = (6 × 10? ³)/2 = 3 × 10? ³ Am²

x = 1/√ (µ? ε? ) = speed ⇒ [x] = [L¹T? ¹]
y = E/B = speed ⇒ [y] = [L¹T? ¹]
z = l/ (RC) = l/τ ⇒ [z] = [L¹T? ¹]
So, x, y, z all have the same dimensions.

$\mathrm{}496.00$

$\mathrm{}PPM=\frac{\text{Mass of Iron}}{\text{Mass of wheat}}\times {10}^6$

Mass of $10=\frac{\text{Mass of Iron}}{100\times {10}^3}\times {10}^6$ $\frac{}{{}^{}}{}^{}$

Mole of $\mathrm{F}\mathrm{e}=1\mathrm{g}\mathrm{m}$ $$

$$ $\mathrm{F}\mathrm{e}=\frac{1}{56}$  contain ${\mathrm{F}\mathrm{e}\mathrm{S}\mathrm{O}}_{4}7{\mathrm{H}}_2\mathrm{O}$ ${}_{}{}_{}$ mole of $$ atom $=1$

$$  $\mathrm{F}\mathrm{e}$ in 1 mole

$\therefore 56\mathrm{g}$  in  $1\mathrm{g}$ $$ mole

$\frac{}{}$  $\frac{1}{56}$ Mass $\therefore$ $$

Area of OABS is A?

Area of SCD is A?
Distance = |A? | + |A? |
A? = (1/2) [13/3 + 1]4 = 32/3
A? = (1/2) × 5/3 × 2 = 5/3
Distance = |A? | + |A? |
= 32/3 + 5/3 = 37/3

Moment of inertia in case (i) is I?
Moment of inertia in case (ii) is I?
I? = 2MR²
I? = (3/2)MR²

T? = 2π√ (I? /Mgd); T? = 2π√ (I? /Mgd)
T? /T? = √ (I? /I? ) = √ (2MR² / (3/2)MR²) = 2/√3

According to the data available on the BTech fees at MPSTME, the total tuition fee at MPSTME is around INR 14 Lacs - INR 25.1 lakh. Candidates may also need to pay other types of fees such as MPSTME hostel fees, institutional fees and other forms of one time payments. Candidates are advised to visit the official website of MPSTME to confirm all the fees payable by them.

$\mathrm{}48.00$

$\mathrm{}\left|w\right|=\frac{1}{2}(6+10)=48\mathrm{J}$

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$\mathrm{}1.52$

$\mathrm{}E=E_0-\frac{0.0591}{4}\mathrm{l}\mathrm{o}\mathrm{g}?{\left. [{\mathrm{H}}^{+}\right]}^4$

$\mathrm{E}=1.23+0.0591\times \mathrm{p}\mathrm{H}$

$E=1.23+0.0591\times \left(5\right)$

$E=1.52$